## blibber 3 years ago Find the matrix B of the linear transformation T(x)=Ax with respect to the basis B = (v1, v2,...,vm): A= [0 1; 2 3] v1= [1; 2] v2= [1; 1]. I can make the [x]_B span...how do you get that final matrix????? (basis change)

1. swissgirl

@Jemurray3 Can u check this out?

2. blibber

To clarify, I can get to B = [ [ 2; 8 ]_B , [ 1; 5 ]_B ] but don't understand how the final B works out to be B = [ 6 4 ; -4 -3 ]

3. Jemurray3

Sorry, just lost a lot of text. Re-typing...

4. Jemurray3

Okay sorry. If I have a vector $\vec{x} = \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right)$ that I want to express as a vector y with respect to the basis vectors v1 and v2, then $\vec{x} = [\vec{v_1} \space \vec{v_2}]\vec{y} = V\vec{y}$ so, $\vec{y} = V^{-1}\vec{x}$ So, the linear transformation $T(x) = A\vec{x}$ can also be written as $T(y) = AV\vec{y}$ But at the end of the day, this quantity $\left( AV\vec{y}\right)$ is a vector with respect to the basis V. To make it a vector with respect to x again, I must apply V inverse: $T(x) = V^{-1}AV\vec{y} = B\vec{y}$ so in conclusion, long story short, $B = V^{-1}AV$

5. Jemurray3

Wait, that last part is wrong. Sorry, ,my explanation got kinda muddled.... the end result is right, but the last two steps weren't explained properly.

6. Jemurray3

Here: If T maps X to X', then $T(x) = \vec{x'} = A\vec{x}$ but $\vec{x'} = V\vec{y'}, \vec{x} = V \vec{y}$ so $V\vec{y'} = AV\vec{y}$ $\vec{y'} = V^{-1}AV\vec{y} = B\vec{y}$

7. blibber

awesome awesome! very cool..thanks sooooo much!! :) one final question (sorry to be annoying..) but so I found V(inv) to be [-2 -4; -2 -5] and then when I go through V(inv)*A*V, I am dealing with [-2 -4; -2 -5] *[0 1; 2 3]*[1 1; 2 1] but I keep ending up with [-10 -6; -12 -7]. I'm pretty sure it's not my matrix multiplication or inverted matrix V that's the problem..do you notice if I'm using a vector out of place somewhere?

8. Jemurray3

V(inv) should be [-1 1 ; 2 -1]

9. Jemurray3

and V should obviously be [1 1 ; 2 1]

10. blibber

OH MY GOODNESS!!!!! haha okay, so I was a little ahead of myself in my notes..my ridic V(inv) above was actually my INCORRECT V(inv)*A already..that's what happens when your work gets all jumbled up, I guess. But my inverted V was wrong anyway..I guess I should take a step back and work on that now hahaha. THANK YOU SO MUCH!!! (exam in 15 hours..angsting so hard right now!) you're awesome! =D

11. blibber

and thanks, swissgirl for hooking me up with a math life-saver! (sorry, im new to the site and dont know how to do anything yet..) im off to study some more!

12. Jemurray3

No problem, best of luck.