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anonymous
 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)
anonymous
 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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Jemurray3 Can u check this out?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To 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 ]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, just lost a lot of text. Retyping...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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, \[B = V^{1}AV \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here: 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 \[ V\vec{y'} = AV\vec{y} \] \[\vec{y'} = V^{1}AV\vec{y} = B\vec{y}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0awesome 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0V(inv) should be [1 1 ; 2 1]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and V should obviously be [1 1 ; 2 1]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OH 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and thanks, swissgirl for hooking me up with a math lifesaver! (sorry, im new to the site and dont know how to do anything yet..) im off to study some more!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No problem, best of luck.
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