Last basis q, I swear

Printable View

Feb 28th 2009, 09:12 PM
smellatron

Last basis q, I swear

Find a basis B (fancy B) of R^2 such that the B-matrix (fancy B) of the linear transformation

T(x) = [-5 -9
4 7]x (Where x is a vector) is B (regular B) = [1 1
0 1]

They give [-9 and [0 as an example of an answer. Just not sure how
6] 1] they get that answer.

http://www.mathhelpforum.com/math-he...c/progress.gif
Mar 1st 2009, 11:20 AM
GaloisTheory1

Quote:

Originally Posted by smellatron

Find a basis B (fancy B) of R^2 such that the B-matrix (fancy B) of the linear transformation

T(x) = [-5 -9
4 7]x (Where x is a vector) is B (regular B) = [1 1
0 1]

They give [-9 and [0 as an example of an answer. Just not sure how
6] 1] they get that answer.

http://www.mathhelpforum.com/math-he...c/progress.gif

opps. i misinterpreted your question. disregard.
Mar 2nd 2009, 05:37 AM
HallsofIvy

Quote:

Originally Posted by smellatron

Find a basis B (fancy B) of R^2 such that the B-matrix (fancy B) of the linear transformation

T(x) = [-5 -9
4 7]x (Where x is a vector) is B (regular B) = [1 1
0 1]

They give [-9 and [0 as an example of an answer. Just not sure how
6] 1] they get that answer.

http://www.mathhelpforum.com/math-he...c/progress.gif

Saying that, with basis {u, v}, T has matrix representiation $begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ means that Tu= u and Tv= u+ v.

So you want to have $\begin{bmatrix}-5 & -9 \\ 4 & 7\end{matrix}\begin{bmatrix}x \\ y\end{bmatrix}= \bmatrix{x_1 \\ y_1}$ and $\begin{bmatrix} -5 & 9 \\ 4 & 7\end{bmatrix}\begin{bmatrix}\begin{bmatrix}x2 \\ y2\end{bmatrix}= \begin{bmatrix}x1+ x2 \\ y1+ y2$ .

That gives you four equations to solve for x1, y1, x2, y2.