i need some explain about the matrix change of basis cuz i want understand it deeply , so anyone help me ?

- dinamix

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- schrodinger

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- dinamix

@amistre64

- dinamix

@ikram002p

- dinamix

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## More answers

- dinamix

@dan815

- dinamix

@IrishBoy123 its not sweet nut now

- Empty

Explain what you know or think about change of basis currently.

- dinamix

@Empty i mean when i have Linear map we use matrix for change of basis , u study this thing in algebra my mind breaks down when i see it , and i dont know why i study this thing

- Empty

Oh ok, well if that's the case then I can explain it to you. :)

- dinamix

@Empty ty

- dinamix

@nincompoop

- dinamix

@Jamierox4ev3r

- Empty

|dw:1440543471583:dw| So here the vectors are not the unit vectors! They are:
\(\vec a = \langle 2,0\rangle\) and \(\vec b = \langle 0,2\rangle\)
So when I write that combination of vectors \(\vec v = \langle 3, 1\rangle \) that means with respect to this basis! Written out a little more: \(\vec v = 3\vec a + 1\vec b\). So really this is shorthand for a linear combination of our basis vectors.
But let's think about this, if we are looking at where to drill holes on a square sheet of metal, that location, vector \(\vec v\) is NOT going to be \( \langle 3, 1\rangle \). What we need to respect is that the vector \(\vec v\) is pointing to an invariant location in space! The numbers are just what we use to do calculation and help us program our drilling machine for example. So due to the precision of drilling, the machine might have a different coordinate system than the one I used with \(\vec a\) and \(\vec b\).
So we can have a new coordinate system for the computer that manages the drilling machine that looks like this: |dw:1440544380250:dw| So here we have the same point, but represented with a different basis. Since we know \(3 \vec a + 1 \vec b = 9 \vec x + 3 \vec y\) this is one kind of made up example where we can solve for these basis vectors.
Really this is a simplified example but I hope it kinda gives the idea of why this might be useful. Try solving for \(\vec x\) or \(\vec y\) maybe or ask any kind of clarifying questions I can try to help! :D

- dinamix

but this u spoke only about vectors , u didnt speak about matrix i know we have relation betwen us

- Astrophysics

So the change is a scaling factor basically?

- dinamix

@Empty now i understand , u give me simple exmple to understand change of basis ty

- Empty

Right, well the matrix follows from this whole discussion here, because we have expressed \(\vec v\) as a linear combination of the vectors \(\vec a\) and \(\vec b\) we can write it and then convert it to matrices, I think that will help!
\[
3\begin{bmatrix}
2\\
0
\end{bmatrix} +1 \begin{bmatrix}
0\\
2
\end{bmatrix}= \begin{bmatrix}
6\\
2
\end{bmatrix}\]
Ahhh we can now write it as the matrix form now!
\[\begin{bmatrix}
2 & 0\\
0 & 2
\end{bmatrix}\begin{bmatrix}
3\\
1
\end{bmatrix} = \begin{bmatrix}
6\\
2
\end{bmatrix}\]
So the matrix represents the basis vectors and the vector multiplying it is the vector with respect to that basis. Now we an similarly solve for what basis has this representation like I showed in my last post:
\[\begin{bmatrix}
x_1 & y_1\\
x_2 & y_2
\end{bmatrix}\begin{bmatrix}
9\\
3
\end{bmatrix} = \begin{bmatrix}
6\\
2
\end{bmatrix}\]
Where does the change of basis come from? We can set these two alternative representations equal to each other and it will look like this:
\[A \vec w = B \vec u = \vec v\] Where A is the basis vectors \(\vec a\) and \(\vec b\) and then B is the basi vectors \(\vec x\) and \(\vec y\) and \(\vec v\) is just the vector (6,2).
Then we are just finding the inverses of these matrices to go between the bases A and B,
\[B^{-1}A \vec w = \vec u\] so the matrix \(B^{-1}A\) converts the representation of our vector \(\vec v\) in as \(\vec w\) in in the A basis to the vector \(\vec u\) in the B basis.

- Empty

I guess I typed all this up but oh well hahaha

- Astrophysics

How do you go from \[3\begin{bmatrix} 2\\ 0 \end{bmatrix} +1 \begin{bmatrix} 0\\ 2 \end{bmatrix}= \begin{bmatrix} 6\\ 2 \end{bmatrix}\] to \[\begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}\begin{bmatrix} 3\\ 1 \end{bmatrix} = \begin{bmatrix} 6\\ 2 \end{bmatrix}\] just addind the two vectors first?

- Astrophysics

adding wouldn't work xD

- Empty

Adding wouldn't work?

- dinamix

its work

- dinamix

i know this thing about matrix

- Empty

@dinamix Good question by the way!

- Empty

Work out this: \[3\begin{bmatrix} 2\\ 0 \end{bmatrix} +1 \begin{bmatrix} 0\\ 2 \end{bmatrix}=\begin{bmatrix} 6\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 2 \end{bmatrix}= \begin{bmatrix} 6\\ 2 \end{bmatrix}\]
Matrix multiply this:
\[\begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}\begin{bmatrix} 3\\ 1 \end{bmatrix} = \begin{bmatrix} 3*2+1*0\\ 3*0+1*2 \end{bmatrix}=\begin{bmatrix} 6\\ 2 \end{bmatrix}\]
I don't know really how to explain this other than this is why matrix multiplication is defined this funky way to begin with.

- Astrophysics

Ooh ok, so it's just another property

- Astrophysics

kind of

- Astrophysics

Ok thanks haha, I was just looking at it and thinking how in the world did you make that but this makes a bit more sense.

- dinamix

astro this things is easy , we are speak about change of basis linear map

- Astrophysics

I understand it, just thought it was a bit weird to from 1 x 1 matrices to 2 x 2 so I questioned it haha, maybe I knew about it before but since empty is already here I thought I'd ask anyways Xd

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