Question

Question below:

Answer 1

Let \[\left\{ u _{1}, u_{2} \right\} \space and \space \left\{ v_{1},v_{2} \right\}\] be ordered basis for R^2 where \[u_{1} = \left(\begin{matrix}1 \\ 1\end{matrix}\right)\] , \[u_{2} = \left(\begin{matrix}-1 \\ 1\end{matrix}\right)\] and \[v_{1} = \left(\begin{matrix}2 \\ 1\end{matrix}\right)\] , \[v_{2} = \left(\begin{matrix}1 \\ 0\end{matrix}\right)\] Let L be the linear transformation defined by \[L(x) = (-x_{1},x_{2})\] and let B be the matrix representing L wrt [u1,u2}

a) Find the transition matrix from change of basis from U to V
b) Find the matrix A representing L wrt V by computing \[S^{-1}AS\]

Answer 2

I've got B to be \[B = \left[\begin{matrix}-1 & 1 \\ 1 & 1\end{matrix}\right]\]

Answer 3

@amistre64 @myininaya @hartnn

Answer 4

@Directrix @Zarkon

Answer 5

Should be \[S^{-1}BS\]

Answer 6

@Hero ?

Answer 7

@terenzreignz ??

Answer 8

First, express \(v_1\) and \(v_2\) as linear combinations of \(u_1\) and \(u_2\)

Answer 9

Wait, topsy turvy, express u1 and u2 at linear combinations of v1 and v2 D:

Answer 10

u1 = v1 - v2
u2 = v1 - 3v2

Answer 11

So that makes s \[S = \left[\begin{matrix}1 & -1 \\ 1 & -3\end{matrix}\right]\]

Answer 12

No, you arrange them vertically XD
\[\Large S^\color{red}T = \left[\begin{matrix}1 & -1 \\ 1 & -3\end{matrix}\right]\]

Answer 13

Oh yeah, i knew that xD

Answer 14

\[S = \left[\begin{matrix}1 & 1 \\ -1 & -3\end{matrix}\right]\]

Answer 15

So \[S^{-1} = -\frac{1}{2}\left[\begin{matrix}-3 & -1 \\ 1 & 1\end{matrix}\right]\]

Answer 16

Okay, sure... (I'll take your word for that)

Answer 17

Yeah its right, i checked :P now i just gotta multiply all this together D:

Answer 18

I got it from here, thank you again for your help! :D

Answer 19

Awesome. No problem :D