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Empty
 one year ago
Is it possible for A^2=0 and A is not equal to 0?
Empty
 one year ago
Is it possible for A^2=0 and A is not equal to 0?

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Empty
 one year ago
Best ResponseYou've already chosen the best response.3I'll admit, I'm being ambiguous on purpose here haha.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[A=\begin{bmatrix}0&\alpha \\0&0\end{bmatrix}\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Alright you win, I can't pull a fast one on you for a nilpotent matrix.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1interesting, is that the only matrix form with degree 2 ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, including the transpose I think so. It's interesting because dual numbers are like the complex number representation of a matrix, only instead of giving us rotations they give us shearing transformations instead when we multiply them.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so they do stuff like transforming a square into rhombus etc is it

Empty
 one year ago
Best ResponseYou've already chosen the best response.3So for contrast to see more vividly: \[a+\epsilon b= a\left[ \begin{array}c 1 & 0\\0 & 1\\\end{array} \right] + b\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] = \left[ \begin{array}c a & b\\0 & a\\\end{array} \right]\] \[a+i b= a\left[ \begin{array}c 1 & 0\\0 & 1\\\end{array} \right] + b\left[ \begin{array}c 0 & 1\\1 & 0\\\end{array} \right] = \left[ \begin{array}c a & b\\b & a\\\end{array} \right]\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I just found this now so I don't really know anything much but you can tell by simply multiplying out the rules: \[(a+b \epsilon)(x+y \epsilon) = ax+(bx+ay)\epsilon\] and the part \(by\) we would normally have is thrown away which is the difference between the rotation and shear! Interesting!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Okay I have a little experience with transformation matrices, so im trying to relate that with these nilpoint matrices Consider a 2 dimensional point on plane \(z=1\) : \((x,y,1)\) \[\begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{bmatrix} a&\color{red}{c}&e\\\color{red}{b}&d&f\\0&0&1\end{bmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\] if i remember correctly those red numbers represent the shear effect

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ahhh perhaps what I've written is confusing for the same reason that if I write out: \[(a+b i)(x+y i) = (axby)+(bx+ay)i\] This represents not only a rotation but a stretching. By the same reasoning the equation below is not only strictly shearing, but also extension at the same time. \[(a+b \epsilon)(x+y \epsilon) = ax+(bx+ay)\epsilon\] I'm essentially using their line her to come to this conclusion from the wikipedia article: http://prntscr.com/7o8mzb Specifically in my mind the only difference is the real part (x axis) of the complex numbers is shifted to the left by \(yb\) while in the dual numbers we aren't shifting left by \(yb\) and are instead solely moving upwards (and stretching in this case) but in that article they do a pure shear I believe.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Specifically I'm interested in grasping the "geometry" section of the article here: https://en.wikipedia.org/wiki/Dual_number#Geometry which should answer our questions about shearing.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Just playing around right now, I noticed there's an interesting form for powers of dual numbers: \[d=a+b \epsilon\] \[d^n = (a+ b \epsilon)^n = a^n + n a^{n1}b \epsilon\] Which is just the first two terms of the binomial theorem. I'm sort of just playing around and trying to figure it out since I don't really understand this but it's interesting and new and might have some interesting uses for solving problems since it's so simple to use.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0don't have an A matrix with 0 entires ...then A^2 won't be equal to 0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0A^2 = A(A) with matrix multiplication

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I'm talking about this matrix specifically: \[\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] \left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] = \left[ \begin{array}c 0 & 0\\0 & 0\\\end{array} \right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought it's not possible. I wasn't thinking about matrices.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1nice thats because \(\epsilon^k=0\) for \(k\gt 1\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.3In effect it seems as though I've really calculated more impressive looking haha: \[\left[ \begin{array}c a & b\\0 & a\\\end{array} \right]^n = a^n\left[ \begin{array}c 1 & \frac{nb}{a}\\0 & 1\\\end{array} \right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What about: \[ \delta = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \implies \delta^2= \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}^2 =\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} = 0 \]And we might think: \[ i = \epsilon  \delta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there any way to compare dual numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually it seems \(\delta = \epsilon^T\), so we would say: \[ i = \epsilon  \epsilon^T \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Interesting one application of dual numbers is that they allow you to do automatic differentiation! Which makes sense I think when you look at them. http://prntscr.com/7o8sqt So specifically I'm guessing because we have \[(a+ b \epsilon)^n = a^n+na^{n1} b \epsilon\] we subtract a^n and divide by epsilon (well whatever that means). Since these are not invertible. I don't know, that's a good question @wio . One thing I'm noticing is we can possibly make a "dual conjugate" which makes the epsilon term negative but what about if we transpose the matrix, that almost seems to correspond to another operation altogether too.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Let's suppose we define \[\epsilon^* = \epsilon \] \[\epsilon ^T\]and allow us to just put a transpose on it to represent the transposed matrix, there's no other thing we will do. Then if we take the conjugate transpose of a dual number and add it to itself we get a complex number.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Example: \[(a+b \epsilon)^* = ab \epsilon \]\[(a+b \epsilon)^T = a+b \epsilon^T\] Combine these operations to make a "Hermitian" operator:\[(a+ \epsilon)^H = ab \epsilon^T\] Now we add: \[(a+b \epsilon) + (a+ b \epsilon)^H = 2a+bi\] Hmm... good idea @wio!!!

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Interesting thing I've just read is by considering the power series of a function it naturally falls out that: \[f(a+b \epsilon) = f(a)+b f'(a) \epsilon\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I wonder if there's a quaternion analog for dual numbers as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \epsilon^T\epsilon = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} \]Therefore \[ \epsilon \epsilon^T +\epsilon^T\epsilon = 1 \]And we can say: \[ a\epsilon\epsilon^T+b\epsilon+c\epsilon^T+d\epsilon^T\epsilon = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \]From \(\epsilon\) and its transpose, you can get a full \(2\times2\) matrix.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Interesting, we can get to all 4 parts by algebra and kind of throwing away the matrix in a sense. Hmmm this seems more fundamental than complex numbers in a sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think there is a single means of generalizing dual numbers. A while back I came up with a way to generalize imaginary numbers that didn't lead to quaternions.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Hmmm well in this light it almost feels as though we have a companion to the "minus sign" and we are using the "transpose" as sort of another minus sign in a different sense here. Specifically when I said the thing about quaternions I meant the Cayleywingspanson construction https://en.wikipedia.org/wiki/Cayley%E2%80%93wingspanson_construction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What happens when you do \(\epsilon^0\)? Are we saying it is \(1\) or indeterminate?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3When I say more fundamental I also mean that we can see that all rotations are really a specific form of combining two shears. Now that we've decoupled the shears into two parts, it seems like maybe we'd be able to rotate along an ellipse or hyperbola if we desire. According to the wikipedia article \(\epsilon^0 = 1\) from calculating it here: http://prntscr.com/7o8yrw

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I am fine with the idea of interpreting it as the identity matrix I think.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I feel as though instead of previously we were just throwing complex numbers into a matrix because it happened to work, now it seems to be crucial since we are now giving each entry of the matrix a specific role to play in our algebra. One shears in one axis, one shears in the other axis, one stretches in one axis and the other stretches in the other axis. At least I believe this is what we can now understand a 2x2 matrix to represent.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Complex numbers are just the case when the shearing is identical and the stretching on the axes is identical which causes circular motion and uniform stretching that commutes with it. I could be wrong on this but this is my hunch, I don't know, this is the direction I'm sorta being dragged along I guess haha weird stuff.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Makes me wonder what happens if you do something like: \[ a+b\alpha = \begin{bmatrix} a+b& b \\ 0&a \end{bmatrix} \]It looks like \[ (b\alpha)^2= \begin{bmatrix} b&b \\ 0&0 \end{bmatrix}^2= \begin{bmatrix} b^2&b ^2\\ 0&0 \end{bmatrix}= b^2\alpha \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But might not be closed under multiplication:

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Interesting yeah. \[(aI+b \alpha)^2 = a^2I+2ab \alpha + b^2 \alpha^2\] We can easily and quickly show \[\alpha^2=\alpha\] so \[(aI+b \alpha)^2 = a^2I+(2ab + b^2) \alpha\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had just computed: \[ (a+b\alpha)(c+d\alpha) = ac+(ad+bc+bd)\alpha \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Another thing that I see them saying is that no dual number has an inverse but you can still do division. I think this is essentially the same idea as clifford algebra division of vectors kind of thing but it doesn't exactly translate to matrices as we know them so I'd like to get a better foundation on this operation: \[\frac{a+b \epsilon}{x+ y \epsilon} = \frac{a+b \epsilon}{x+ y \epsilon} \frac{x y \epsilon}{x y \epsilon} = \frac{a}{x} + \frac{ bxay}{x^2} \epsilon\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ (a+b\alpha)^n = \sum_{k=0}^n{n\choose k}a^{nk}(b\alpha)^k = a^n\alpha^0+\sum_{k=1}^n{n\choose k}a^{nk}b^k\alpha \]Not sure what \(\alpha^0\) would look like.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I think in general we can say that for matrices \(A^0 = I\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.3As long as A isn't the zero matrix. The reasoning for why that would be a suitable definition is just by how we define the exponent rules: \(A^{n+m} = A^nA^m\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, that does make sense here: \[ (a+b\alpha)^n = a^n + \left(\sum_{k=1}^n{n\choose k}a^{nk}b^k\right)\alpha \]Seems similar to what you got for \(n=2\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you define modulus or magnitude for dual numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess it's really asking what your inner product would be.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think that imaginary numbers would use the determinate for the inner product.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3To find out what an inverse dual matrix representation would be, I calculated out multiplying a matrix with values and plugged it in to get: \[\left[ \begin{array}c \frac{1}{x} & y\\0 & \frac{1}{x}\\\end{array} \right]\] is the inverse of \[\left[ \begin{array}c x & y\\0 & x\\\end{array} \right]\] Which supposedly doesn't exist but now I'm curious if multiplying these gives us 1. Inner product is a good question, I don't know. Since we can turn a dual into a complex number we might be able to just hijack that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But for dual numbers, the determinate seems to give \(\langle a+b\epsilon, a+b\epsilon\rangle = a^2\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the alpha concept, it would seem determinate inner product would give \[ \langle a+b\alpha, a+b\alpha\rangle = a^2+ab \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I guess I don't know if I even understand these things geometrically yet in the "dual plane". While I can visualize complex number multiplication I can't quite visualize dual multiplication yet, so I guess that's what's to look into now for me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm \[(a+b\alpha)^{1}=\begin{bmatrix} a+b&b\\0&a \end{bmatrix}^{1} =\frac{1}{a^2+ab}\begin{bmatrix} a&b\\0&a+b \end{bmatrix}^{1} =\frac{(a+b)+b\alpha}{a^2+ab} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3So while \(e^{i t} = \cos t + i \sin t\) it appears that we have \(e^{\epsilon t} = 1+ \epsilon t\) which is supposedly some kind of parabolic rotation. I'm not sure, I'm just about to download and skim this paper: http://arxiv.org/abs/0707.4024

Empty
 one year ago
Best ResponseYou've already chosen the best response.3See @wio I said earlier I was studying probability and now I'm onto this already haha

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Interesting, another alternative is that we can use a different matrix other than i or epsilon to create a hyperbola.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{\epsilon t} = \sum_{n=0}^\infty \frac{(\epsilon t)^n}{n!} = 1 + \epsilon t + \sum_{n=2}^\infty \frac{(\epsilon t)^n}{n!} \]Where \[ \sum_{n=2}^\infty \frac{(\epsilon t)^n}{n!} = \sum_{k=0}^\infty \frac{(\epsilon t)^{k+2}}{(k+2)!} = \sum_{k=0}^\infty \frac{0\cdot (\epsilon t)^k}{(k+2)!} = 0 \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3All of the shapes traced out by the "rotations" are defined by this single function: \[x^2i^2y^2=1\] So if i is the traditional imaginary number we have \[x^2+y^2=1\] if i is our new dual epsilon we have: \[x^2=1\] and if we have this other alternative hyperbolic i we have \[x^2y^2=1\] I wonder what matrix they're using for this hyperbolic i that squares to 1 but isn't 1.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Probably \[\left[ \begin{array}c 0 & 1\\1 & 0\\\end{array} \right]\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3This is illuminating: http://prntscr.com/7o9ctw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One way to generalize dual numbers: \[ \epsilon_3 = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}, (\epsilon_3)^2 = \begin{bmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \]We have: \[ (\epsilon_3)(\epsilon_3)= (\epsilon_3)^2 \\ (\epsilon_3)^3 = (\epsilon_3)^2(\epsilon_3) = 0 \]This is similar to my previous generalization of imaginary numbers, where tried to start with an identity: \[ (i_3)^3 = 1 \]It turned out that \(i_3\) was just a 6th of a rotation around the unit circle, if you projected onto the imaginary plane.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0However, the 'quaternion' approach to it might be more along the lines of: \[ \epsilon = \begin{bmatrix} 0&0&1&1 \\ 0&0&1&1 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} \]Or \[ \epsilon = \begin{bmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3The amazing parabolic sine and cosine functions lol, and they satisfy the hilarious \[e^{\epsilon t} = cosp(t) + \epsilon sinp(t) = 1+\epsilon t\] \[sinp(x) = x=\frac{e^{\epsilon x}e^{\epsilon x}}{2 \epsilon} \]\[cosp(x)=1=\frac{e^{\epsilon x}+e^{\epsilon x}}{2 }\] \[\frac{d}{dx} sinp(x) = cosp(x)\]\[\frac{d}{dx} cosp(x) = 0\] So at least it feels natural in a stupid way.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ah interesting I see what you're saying. I think I am sort of seeing how pauli spin matrices might just be how we write quaternions using matrices and complex numbers or something, I'm not sure.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Out of curiosity can we extend Fermat's Last theorem to matrices? \[A^n+B^n=C^n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{\alpha t} = \sum_{n=0}^\infty \frac{(\alpha t)^n}{n!} = \sum_{n=0}^\infty \frac{\alpha^n t^n}{n!} = 1+ \sum_{n=1}^\infty \frac{\alpha t^n}{n!} \]We can can start that \(n\) back at zero by adding and subtracting \(\alpha\):\[ 1+ \sum_{n=1}^\infty \frac{\alpha t^n}{n!} + \alpha t^0  \alpha t^0 = 1  \alpha + \alpha\sum_{n=0}^\infty \frac{t^n}{n!} = 1\alpha + \alpha e^{t} \]In summary:\[ e^{\alpha t} =1\alpha + \alpha e^{t} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3This is very illuminating to me since the difference between these three matrices is simply the bottom left value being 1, 0 or 1. http://prntscr.com/7o9h0g It almost seems like we should consider all of these simultaneously with a value in the bottom left that we can change spaces with.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which could also be said as: \[ e^{\alpha t} = 1 + (e^t1)\alpha \]So we could say: \[ \cos\alpha(t) = 1 \\ \sin\alpha(t) = e^t1 \\ \]Very weird result.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, not completely sure about that last one but

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\[\left[ \begin{array}c 0 & 1\\s & 0\\\end{array} \right]^2 = s\left[ \begin{array}c 1& 0\\0 & 1\\\end{array} \right]\] So we have this matrix where the bottom left entry is the value of the square. 0, 1, or 1. How are you coming to that result wio?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am just using the weird identity: \[ e^{\alpha t} = \cos_\alpha(t) + \alpha \sin_\alpha(t) = 1+(e^t1)\alpha \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Though I'm not sure what the underlying assumptions of that identity are.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's more of a definition though, so...

Empty
 one year ago
Best ResponseYou've already chosen the best response.3well as long as you define sine and cosine by this structure: \[sinp(x) = x=\frac{e^{\epsilon x}e^{\epsilon x}}{2 \epsilon} \]\[cosp(x)=1=\frac{e^{\epsilon x}+e^{\epsilon x}}{2 }\] I think you will in a good position.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I'll try to play around with the alpha values a bit to see if I get to it as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That structure? Why?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3It produces an even and an odd function that add to the exponential function because: \[f(x) = \frac{f(x)+f(x)}{2}+\frac{f(x)f(x)}{2}=g(x)+h(x)\] and we have the identities that g(x)=g(x) and h(x)=h(x) like we expect from a sine or cosine.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Also I'm noticing we might be able to generalize your matrix a bit: \[\left[ \begin{array}c s & 1\\0 & 0\\\end{array} \right]^2=s\left[ \begin{array}c s & 1\\0 & 0\\\end{array} \right]\] or in other words \[\alpha_s^2=s \alpha_s\] So you have some control over what the square is, you can make s=0, 1, or 1 for different interesting but also easier to track properties. Something to play with.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's see \[ e^{\alpha t} = 1 + (e^{t}1)\alpha \]So \[ e^{\alpha t}+e^{\alpha t} = 2 + \alpha\bigg(e^t1 + e^{t}1\bigg) = 2+\alpha (e^t+e^{t}2) \]So we have: \[ \frac{e^{\alpha t}+e^{\alpha t} }{2} = 1+\cosh(t) \]For the other one: \[ e^{\alpha t} e^{\alpha t} =\alpha\bigg(e^t  e^{t}\bigg) = \]So \[ \frac{e^{\alpha t} e^{\alpha t}}{2\alpha} = \sinh(t) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm, the only issue thought is that I know division by \(\alpha\) is problematic, just as it is for \(\epsilon\).

Empty
 one year ago
Best ResponseYou've already chosen the best response.3It is weird I'll agree with you hmm.\[\frac{a+b \alpha}{x+ y \alpha} = \frac{a+b \alpha}{x+ y \alpha} \frac{ab \alpha}{x y \alpha} = \frac{ax}{x^2y^2} + \alpha \frac{bxayby}{x^2y^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For \(a+b\epsilon\) we have division problems at \(a=0\) (regardless of \(b\)), and for \(a+b\alpha\) we have division problems at \(a=b\). You could say that for complex we have division problems at \(a^2+b^2=0\) for obvious reasons.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, I am not sure how I feel about "dividing by a matrix" but somehow it seems to just sorta work even though it definitely has problems going on.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think we have to say that at these places, the magnitude ought to be defined as \(0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Everything we have done so far can be generalized as "Using a \(2\times 2\) matrix to define multiplication for a \(2\) component vector."

Empty
 one year ago
Best ResponseYou've already chosen the best response.3But I think we have some interesting relationship that the matrices satisfy a relation like this: \[A^2=sA\] or \[A^2=sI\] So there's some sort of structure here. Hmmm. It's sorta all nonsense in a way though unless we can get something out of it I think.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We could even do Hadamard matrix for this: \[ a+bH = \begin{bmatrix} a+b & a+b \\ a+b & ab \end{bmatrix} \]It appears that \(H^2 = 2I\), so: \[ (a+bH )(c+dH ) = (ac+2bd)+(ad+bc)H \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That one quantum computing tutorial said something about Unitary matrices or something being special.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Oh interesting. I am sort of getting a new idea from this. Yeah a unitary matrix is basically just the complex version of an orthogonal matrix. An orthogonal matrix is the inverse of its transpose while a unitary matrix is the inverse of its hermitian, which is the conjugate transpose. The DFT uses this for instance.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\[M=\left[ \begin{array}c a & b\\c & d\\\end{array} \right]=a\left[ \begin{array}c 1 & 0\\0 & 0\\\end{array} \right]+b\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right]+c\left[ \begin{array}c 0 & 0\\1 & 0\\\end{array} \right]+d\left[ \begin{array}c 0 & 0\\0 & 1\\\end{array} \right]\] \[M=aA+bB+cC+dD\] So now if I square this I'm going to get a more in depth understanding of how each part influences the square rather than looking at our specific cases like we've been doing.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\[M= \left[ \begin{array}c A & B & C & D\\\end{array} \right]\left[ \begin{array}c a\\ b\\ c\\ d\\\end{array} \right]\] So now I've got a matrix of matrices.... Maybe this isn't the right path to take on this haha. Nevermind.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ H^n = \begin{cases} 2^{k}I & n=2k\\ 2^{k}H & n=2k+1 \end{cases} \]I think the power of \(n\) is important, since we can use it in the MacLaurin series of \(e^x\).

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah I think you're right about that. Splitting it up like you've done is also going to make it nice for thinking about the sine and cosine functions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{tH} = \sum_{n=0}^\infty \frac{(tH)^n}{n!} = \sum_{k=0}^\infty \frac{t^{2k}}{(2k)!} + H\sum_{k=0}^\infty \frac{t^{2k+1}}{(2k+1)!} \]Hmmm, not quite trignometric, since we don't have the \(1\) coefficient.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wow. \[ e^{tH} = \cosh(t) + H\sinh(t) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean the fact that \(\cos_H (t) = \cosh\). Seems really coincidental.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Haha weird, the ghost of Hadamard knew his name would coincide with Hyperbola or something XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so here is my opinion. If you have some algebra: \[ a+bA \]where \(A\) is some \(2\times2\) matrix, then what if we define it so that: \[ a+bA = re^{A\theta} = r\bigg(\cos_A(\theta) + A\sin_A(\theta)\bigg) \]Somehow this makes sent to me because \(\theta\) is an angle and \(r\) is a magnitude. I suppose there are still some issues though, because multiplication of \(a+bA\) won't always give you some sort of rotational result. Maybe this pattern is only worthwhile for certain cases like complex numbers.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3No I think you're right, I was thinking something along the same lines, that for every matrix we essentially can define a pair of sine and cosine functions. I am just wondering if we can perhaps extend this to nxn matrices and instead of creating even and odd functions (even is the 2 of 2x2 matrices) so we can make other like "mod 3" functions for our 3x3 matrices for instance.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0You guys should publish a novel on this lol xD

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Haha we just have to publish, writing the novel we already did it lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyway, for rules mentioned before: \[ A^2=cA\implies A^n = c^{n}A \\ A^2 = cI \implies A^n = \begin{cases} c^kI &n=2k \\ c^kH &n=2k+1 \end{cases} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3slight correction:\[ A^2=cA\implies A^n = c^{n1}A \\ \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, that looks right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For that first case \[ e^{At} = \sum_{n=0}^\infty \frac{(At)^n}{n!} =A\sum_{n=0}^\infty \frac{c^{n1}t^n}{n!} = \frac{A}{c}e^{ct} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the second case, I think I'm getting: \[ e^{At} = \cosh(ct) + A\sinh(ct) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, the first case I think I was a bit too hand wavy though... let me double check.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Now that I'm thinking about these generalizations I thought why not make the matrix square to a different matrix altogether? But then I thought, what about if we use group theory? Let's instead just use group theory which already has multiplication defined and allows us to do the same exact things, but we're no longer confined to what a matrix multiplication gives us.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you find matrix such that: \[ A\neq I \land A^3=I \]I'm not sure, but my gut feeling is we need at least a \(3\times3\) to actually get that.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so we have to find... a matrix that makes this true so we can't have Matrix A equal to an identity and have A^3 = to an identity matrix

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so A can't be an identity matrix but using matrix multiplication A(A)(A) that result should come up to an identity matrix

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah good point, just work it out with variables and solve the system of equations... but who wants to do that? haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We're talking about a very ugly system of equations.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3True, but we might be able to simplify it. I think tensor notation makes each equation simpler to look at in a sense. \[A^i_jA^j_kA^k_l = \delta ^i_l\] Or in terms of individual components (in linear algebra terms) where n is the row and m is the column of the entry: \[a_{n1}a_{11}a_{1m}+a_{n1}a_{12}a_{2m}+a_{n2}a_{21}a_{1m}+a_{n2}a_{22}a_{2m}=\delta_{nm}\] So this represents our 4 equations. Plug in n and m equal to 1 or 2 to get them.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[A=\begin{bmatrix}1&0 \\0&e^{i5 \pi/6}\end{bmatrix}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can easily find a solution for \(3\times 3\) though. \[ A = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{bmatrix} \]Then \[ A^3=I \]And I think: \[ A_c = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ c&0&0 \end{bmatrix} \implies (A_c)^3 = cI \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 In our context, the matrix you have provided technically would be something along the lines of a \(4\times 4\), once expanded to real numbers.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ohk I see, I'm not able to follow the conversation completely as it is way too above my head lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1does this work http://www.wolframalpha.com/input/?i=%7B%7B1%2C2%7D%2C%7B3%2F2%2C2%7D%7D%5E3+

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, that could work.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ahhh how did you think up either of these matrices, I'm impressed @ganeshie8 !

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I don't think I let wolfram think lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That matrix would seem to work like: \[ A^n = \begin{cases} I &n=3k \\ A &n=3k+1 \\ A^2 &n=3k+2\end{cases} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe ganeshie can help identify the MacLaurin series we get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{At} = \sum_{k=0}^\infty \frac{t^{3k}}{(3k)!}+A\sum_{k=0}^\infty \frac{t^{3k+1}}{(3k+1)!}+A^2\sum_{k=0}^\infty \frac{t^{3k+2}}{(3k+2)!} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=sum+t%5E%7B3n%7D%2F%283n%29%21++for+n+%3D+0+to+infty For the first one. This looks messy.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\[e^{At}=f(t)+A g(t)+ A^2 h(t)\] where we have \(\omega = e^{i 2 \pi /3}\) and analgously we have our ENTIRELY REAL FUNCTIONS satisfying these "even odd" mod 3 symmetries: \[f(t)=f( \omega t) = f( \omega ^2 t) \\ g(t)=\omega g( \omega t) = \omega^2 g(\omega^2 t) \\ h(t) = \omega^2 h(\omega t) = \omega h (\omega^2 t)\] And additionally we can represent: \[f(t) = \frac{e^{At}+e^{\omega At} + e^{\omega^2 A t}}{3}\] and we have two more corresponding ones, where when we divide by 3 in the denominator we add A and A^2 to match just like we would for complex sine we have to add an i.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Maybe I'm off a bit, since I believe f'(t)=g(t) and g'(t)=h(t) and h'(t)=f(t), but I think what I've said is mostly correct, I am sorta just saying these from memory I didn't actually calculate any of this right now, but I've done this particular thing before.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the link I gave will give you \(f(t)\), and it's messy.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure though if there is a unique matrix solution to the equations anymore, though.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Now I feel like we should just throw away matrices and use group theory since the entries of the matrix are starting to hinder us when we really aren't using them for anything.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But without matrices, it's hard to check out results.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Matrices allow easy definitions of inverses.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Groups allow us to arbitrarily define inverses and we can insure that it's closed simultaneously if we want

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \(A^3 = 1\), then how would you even go about \(A^{1}\)?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3It's its own inverse in that case. dw:1435918078087:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're saying \(A^2 = A^{1}\)?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3No sorry, here so from this cayley table dw:1435918188833:dw I should have just drawn out the whole table. In order to check that every element has a unique inverse we simply check to see that there is only one identity element per column and per row.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think there is a good chance that you can find a matrix which represents any sort of variable you come up with.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I don't doubt it, but I don't think we particularly gain anything from playing with elements in the matrices except longer calculations. Plus it looks like we have multiple different types of matrices that end up obeying the same rules, so it doesn't feel very unique and this shifts the focus a bit I guess. I don't know, just something to think about.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3We now have the ability to define noncommutative structures and other larger weirder things I think while it still being manageable since we're looking it up in a chart rather than carrying out multiplication on nxn matrices which becomes difficult for n=3 and higher. Suppose we want \(A^{8}=I\) I think we would be in trouble.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, to be really generalized for the 2 component case, we can define: \[ A^2 = aI + bA \]For the \(n\) component case , we can define: \[ \bigg( \sum_{k=0}^{n1} a_kA^k \bigg)\bigg( \sum_{k=0}^{n1} b_kA^k \bigg)=\bigg( \sum_{k=0}^{n1} c_kA^k \bigg) \]For coefficients as vectors \(\mathbf a\) and \(\mathbf b\) we say \(\mathbf c =f(\mathbf a, \mathbf b)\).

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ultimately it seems like we're really just using these matrices to define how we multiply "complexish numbers" in our vector space and plugging them into power series to see if we get a cool set of "rotation" functions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I suppose there might be some kinds of \(f\) which can't be achieved through matrix multiplication.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I'm sorta not sure how to raise A to the third power from this definition: \(A^2 = aI+bA \implies A^3 = ?\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.3The convolution you've written there is interesting to me though, I would like to understand this more I don't quite know what you're saying but I think I like where it's going.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ A^3 = aA + b(aI+bA) = ab + (a+b^2)A \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to generalize, but hmmm

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Interesting your definition of A^2 means we can write: \(A= \frac{1}{b}A^2  \frac{a}{b}I\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, the most general I would ever want to get is to say we're looking for multiplication functions, which can be described as \(f:\mathbb R^d \times \mathbb R^d \to \mathbb R^d\). Obviously that will keep things closed.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But for \(d=2\), and working in a vector space, the only thing we really need to define is \(A^2\) because algebra will solve the rest.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Sounds good to me. I feel like from this perspective we can define multiplication between any of our "complexish" numbers to be anything from rotations, to shear, to whatever and then we can just go ahead and define the group theory elements to match what we want.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @wio \[ A^3 = aA + b(aI+bA) = ab + (a+b^2)A \] \(\color{blue}{\text{End of Quote}}\) For complex numbers, we would say \((a,b) = (1,0)\), and so: \[ i^3 = (1)(0)+((1)+(0)^2)i = i \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3That way we can simply calculate stuff with algebra knowing it obeys our rules.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For dual numbers \((a,b) = (0,0)\), and so: \[ \epsilon^3 = 0 \]For \(\alpha\), we'd say \((a,b) = (0,1)\), and we get: \[ \alpha^3 = (0)(1) + ((0)+(1)^1)\alpha = \alpha \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we consider only cases where \((a,b) \in \{1,0,1\}^2\), that gives us about \(9\) combinations we can mess with.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Hmmm I'm trying to find some interesting multiplication definition.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using: \[ A^2 = a+bA \]as a generalization, then: \[ (c+dA)(f+gA) = (cf+adg) + (cg+df+bdg)A \]Hmmm, it looks overly general, but I've been thinking something...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we want to do \[ \frac{c+dA}{f+gA} \]All we need is to find a way to get rid of that \(A\) in the denominator.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the only way this can be fun is if we have something we want to try to find.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We have found \(\sin_i\) and \(\cos_i\) for the different conic sections, so that is off the table.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ellipse or a torus is what I've been searching for, but I think I am too tired to think up something that'll do that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But those are 3d objects, not conic sections

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And ellipse corresponds to original trig functions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, hmmm, I suppose circle is a type of ellipse so...

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I think an ellipse might be nice to be able to have operations that let us rotate the ellipse itself like this dw:1435921176282:dw Also what's stopping us from doing more than one? We can have something like quaternions with multiple parts interacting.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3One idea is that whatever manages the ellipses will collapse down into complex numbers

Empty
 one year ago
Best ResponseYou've already chosen the best response.3But ellipses can stretch and rotate, something circles can't do. I mean you can rotate a circle, but it's meaningless.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, now that I think about it... \(i\) corresponds to the circle, a class of an ellipse. \(\epsilon\) corresponds to a specific class of parabola. And then \(H\) corresponded to a class of hyperbola.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And \(H\) would be like \((2,0)\), I think. That is: \[ H^2 = (2) + (0)H = 2 \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, actually since these are all sorta just separate sections of a conic maybe we can do a linear combination of these to get an ellipse?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So clearly we are getting conic sections for \((x,0)\) configurations.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I never really even bothered to try to calculate something like: \[(x+yi+z \epsilon)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Conic sections have a certain property, eccentricity or something.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah I don't know either haha, I just know that elliptic integrals are hard and finding arclength, there's a possibility that what we're doing gets us something interesting to play with there.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For a circle, we use \((1,0)\). For a hyperbola we use \((2,0)\). For a parabola we use \((0,0)\). Maybe it is just \((x1,0)\) where \(x\) is the eccentricity. Just a hypothesis.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3That sounds like a great idea to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Eccentricity of an ellipse is \(\sqrt{1b^2/a^2}\).

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I'm too tired to really learn anything new right now but it'll give me something to think about later

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, maybe another day we can experiment.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Come up with some way to test hypothesis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And perhaps figure out how \((0,y)\) changes things up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's clear that \(\alpha\) gave us some clear insight into how that might work.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah definitely, I don't think we should throw away matrices they helped us quite a bit in concrete ways of determining what products _should_ be. For instance right now I'm just calculating \(\epsilon * i \) and \(i*\epsilon \) as something to explore with my brain later lol.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3In some sense we're sorta thinking in terms of instead of functions of a complex variable, more like functions of a 2x2 matrix variable.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looking at the eccentricity of a hyperbola, I think my hypothesis was wrong. Hadamard seemed to give us a standard hyperbola, so its eccentricity should be: \[ \sqrt{1+\frac {1^2}{1^2}} = \sqrt{2} \]And \(\sqrt{2} + 1 \neq 2\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm going to bed, but now I have a goal.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3In my mind I'm imagining the conic section. If you cut it one way you get a circle and if you cut it the other way you get a hyperbola. I'd like to rotate that plane that we're cutting through to somewhere inbetween there. Actually... I think I know how to do it, because earlier we had: \[\left[ \begin{array}c 0 & 1\\s & 0\\\end{array} \right]\] where s=1 representing a hyperbola, s=0 representing a parabola, and s=1 representing a circle. And I know that this is like the same spacing, so if 1<s<0 is what we use, we will be on an ellipse I think.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's clear to me that the matrix you have presented corresponds to \((s,0)\). Hmm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there something to correspond to \((0, t)\), now I wonder...

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Haha well perhaps we have chosen the transpose arbitrarily, let's just choose the lower triangular version of this maybe?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess it would be going back to: \[ A^2 = tA \]We know for \(t=1\) we can use: \[ \begin{bmatrix}1&1\\0&0\end{bmatrix} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since that is what was used for \(\alpha\).

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Woah absolutely fascinating. Instead of calling them real and imaginary parts they call it bosonic and fermionic directions for real and \(\epsilon\) directions. This is used for particle physics. http://prntscr.com/7ob3pm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I found it:\[ A = \begin{bmatrix} 1&1 \\ t1&t1 \end{bmatrix} \implies A^2 = tA \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0This is probably the longest thread I've ever been on xD!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wonder if there is a matrix such that \[ A^2 = sI + tA \]If we can find such a matrix, then matrix would be just fine.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's kinda funny, because ultimately what we have here is: \[ A^2 tAsI \]Which is like a quadratic equation for matrices.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is an ugly solution, but: \[ A = \bigg(\frac{t+\sqrt{t^2+4s}}{2} \bigg) I \implies A^2 = sI +tA \]But since \(A=cI\), it sort of doesn't count. I wonder if there is another solution.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the reason it lets you do automatic differentiation is because dual numbers capture a property of nilpotency used by Fermat, Newton, etc. that \(\varepsilon^2=0\) (recall Fermat's https://en.wikipedia.org/wiki/Adequality ); this is used in functional languages to explicitly extend real computational functions to dual numbers and 'compute' their derivatives in a way unlike symbolic or numerical differentiation. it's mainly used in machine learning contexts to come up with derivatives of complex functions (like a feedforward neural net) for error propagation in learning algorithms fyi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$f(a+b\varepsilon)=f(a)+\varepsilon f'(a)+\frac12\varepsilon^2 f''(a)+\dots=f(a)+\varepsilon f'(a)$$this is kinda the same idea behind the trick used in numerical differentiators to evaluate derivatives of \(x\) at \(x+i\) for smaller error terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ps the 'hyperbolic' numbers you were talking about are just the splitcomplex numbers with \(j^2=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is the most straightforward way of encoding splitcomplex numbers in \(M_2(\mathbb{R})\) https://en.wikipedia.org/wiki/Splitcomplex_number#Matrix_representations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and as you noticed \(j^2=1\) ends up causing \(a+bj=1\) to giv ethe unit hyperbola (a^2b^2=j\) and its paramterized by \(\exp(bj)=\cosh(b)+j\sinh(b)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the problem with an ellipse is that there is no 'unit' ellipse, but we can define an ellipse as \(a^2+k^2b^2=1\) which just reduces to \(a+bw\) where \(w\) is some scaled kind of \(i\) (specifically \(w=ki\)), so it ends up giving rise to the same algebra \(\mathbb{C}\) as just \(i\) alone  boring. let's rehash the different types of conics: 1) ellipses (using \(i^2=1\)) 2) parabolas (using \(\varepsilon^2=0\)) 3) hyperbolas (using \(j^2=1\)) ... note the correspondence which is actually a consequence of the limitations we face when trying to come up with twodimensional unital algebras over the reals https://en.wikipedia.org/wiki/Hypercomplex_number#Twodimensional_real_algebras

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\(\color{blue}{\text{Originally Posted by}}\) @wio I wonder if there is a matrix such that \[ A^2 = sI + tA \]If we can find such a matrix, then matrix would be just fine. \(\color{blue}{\text{End of Quote}}\) Yes because the CayleyHamilton theorem says every square matrix satisfies its own characteristic equation. That is to say we can replace A with \(\lambda\) which represent the eigenvalues of A and solve. \[\lambda^2 = s+t \lambda\] The solution to this quadratic has two roots so we can throw it into a 2x2 diagonal matrix: \[\left[ \begin{array}c \lambda_+ & 0\\0 & \lambda_\\\end{array} \right]\] So this is a nice trick for making symmetric, diagonal matrices that obey any polynomial equation.
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