brinethery
Find a basis for the space V of all skewsymmetric 3x3 matrices. What is the dimension of V?



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brinethery
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I'll be back in a sec. Brewing some coffee. Would you like some?

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Yeah, I will in maybe two hours =P

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Well you need to remember that a basis for any vectorial space generates it and it's vectors are linear independent.

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And 3x3 skewsymmetric matrix is something like this:

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\[A=A^T\]

brinethery
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:)

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Hmm thinking...

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Do you have any idea to begin with?

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Mm you need three matrices am I right?

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Well the dimension is 3 haha

brinethery
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I am stupid haha

brinethery
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and I need a longer explanation b/c A=A^T isn't gonna cut it

brinethery
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soy estupido haha

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uff
I almost burn my brain hehe

brinethery
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I told you linear algebra's no good!

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Well I think that the three vectors are \[\left[\begin{matrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{matrix}\right]\]

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Hmmm you're right

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Is the idea of a basis clear to you?

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Like you know why (1,0,0), (0,1,0),(0,0,1) is the basis of R^3?

brinethery
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yes.

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Well then you know that the basic idea behind a basis is that you can get any vector of the space by forming a linear combination.

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like \[\vec{v}=a\hat{i} + b\hat{j} + c\hat{k}\]

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dw:1337293630890:dw

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well that is the idea in a graphic form.

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If you want to obtain a basis for the 3x3 skew symmet

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ric matrices you need to know that is the general form of matrix of that kind.

brinethery
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okay, I'm following what you're saying

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the definition says that the elements of an skew symmetric matrix are such as: \[a_{ij}=a_{ji}\]

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It seems that that doesn't tell us a lot about those matrices, but you can figure out what you have on the diagonal.

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if i = j, then \[a_{ii}=a_{ji}\]

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which is true if the diagonal is made up zeros.

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and I think that's all. An 3x3 skew symmetrical matrix has this form:


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\[\left[\begin{matrix}0 & a_{12} & a_{13} \\ a_{12} & 0 & a_{23}\\ a_{13} & a_{23} & a_{33} \end{matrix}\right]\]

brinethery
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Looks like it's all zeros down the diagonal.

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shame on my haha

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yeah a_33 = 0

brinethery
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you're a bad math teacher haha! :p

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yeah! I'm terrible =(

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but it's fun!

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well now you can split that matrix on three parts:

brinethery
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Yep this part is where I need explaining...

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\left[\begin{matrix}0 & a_{12} & 0 & \\ a_{12} & 0 & 0\\ 0&0&0\end{matrix}\right]

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\[\left[\begin{matrix}0 & 0 & a_{13} & \\ 0 & 0 & 0\\ a_{13}&0&0\end{matrix}\right]\]

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\[\left[\begin{matrix}0 & 0 & 0 & \\ 0 & 0 & a_{23}\\ 0&a_{23}&0\end{matrix}\right]\]

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Is it clear why I did that?

brinethery
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Oh I see. So each part of the basis will be a 3x3 matrix?

brinethery
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And there will be (3) 3x3 matrices

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yeah! it has to bee a 3x3 matrix

brinethery
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I asked on cramster and some (puta) was really insulting to me AND she gave me the wrong answer.

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what is cramster? and puta?

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we are not finished yet.

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you can do this:

brinethery
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I thought it meant b)it(ch but I guess it's mexican slang for prostitute or whore

brinethery
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but you get the idea.

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\[\left[\begin{matrix}0& 1 & 0 \\ 1 & 0 &0\\ 0&0&0\end{matrix}\right]\]


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ahh I thought I did haha.

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\[\left[\begin{matrix}0& 0 & 1 \\ 0 & 0 &0\\ 1&0&0\end{matrix}\right]\]

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and \left[\begin{matrix}0& 0 & 0 \\ 0 & 0 &1\\ 0&1&0\end{matrix}\right]

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You multiply by the scalars a,b,c each of those matrices and form a linear combination.

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and the result is a skew symmetrical matrix

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I'm sorry If my explanation was awful, and my english worst =P

brinethery
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So the dimension of all skewsymmetric matrices is 2?

brinethery
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or that's only the dimension of the basis?

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the dimesion is the number of vectors of the basis. so it's 3.

brinethery
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hahaha I am so stupid!

brinethery
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WOW. I can't believe I asked that.

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if you have an 4x4 skew symmetrical matrix the dimension of the basis is 4.

brinethery
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:) yeah yeah yeah

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Don't worry, don't judge yourself so severely.

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Do you have any doubt regarding this brinethery?

brinethery
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I have another question for you!
No I don't have any doubt. You did a great job explaining.

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Ok

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Linear Algebra?