1. ## A diagonal basis.

I have the Transformation T s.t:

$\displaystyle T(-2, 2, -1)=(-1, -1,-3)$ $\displaystyle T(0, 2 ,2)=(4, -2, 2)$ and $\displaystyle T(0, 1,-1) = (0,1, -1)$

Find a basis B of R3 such that M_B(T), the B-matrix of T, is diagonal.

I really dont know how to approach this i know i want to find

P so that $\displaystyle M_B(F) = P^{-1}AP$ but how do i find A etC?

2. Originally Posted by Scopur
I have the Transformation T s.t:

$\displaystyle T(-2, 2, -1)=(-1, -1,-3)$ $\displaystyle T(0, 2 ,2)=(4, -2, 2)$ and $\displaystyle T(0, 1,-1) = (0,1, -1)$

Find a basis B of R3 such that M_B(T), the B-matrix of T, is diagonal.

I really dont know how to approach this i know i want to find

P so that $\displaystyle M_B(F) = P^{-1}AP$ but how do i find A etC?
Let $\displaystyle \bold{v}_1 = \begin{bmatrix}-2&2&-1\end{bmatrix}^T, \bold{v}_2 = \begin{bmatrix}0&2&2\end{bmatrix}^T, \bold{v}_3=\begin{bmatrix}0&1&-1\end{bmatrix}^T$
Let $\displaystyle V = \{\bold{v}_1,\bold{v}_2,\bold{v}_3\}$ then $\displaystyle [T]_V = [ T(\bold{v}_1)| T(\bold{v}_2)| T(\bold{v}_3) ] = \begin{bmatrix} -1 & 4 & 0 \\ -1&-2&1\\-3&2&-1 \end{bmatrix} = A$.

Let $\displaystyle B = \{\bold{b}_1,\bold{b}_2,\bold{b}_3\}$. If this basis diagnolizes $\displaystyle T$ then it means $\displaystyle [T]_B = [T(\bold{b_1})|T(\bold{b}_2)|T(\bold{b}_3)]$ is a diagnol matrix.

Let $\displaystyle P$ be the transition matrix from $\displaystyle V$ to $\displaystyle B$ then it means $\displaystyle [\bold{x}]_B = P[\bold{x}]_V$ where $\displaystyle [\bold{x}]_B$ is coordinate vector with respect to $\displaystyle B$ and $\displaystyle [\bold{x}]_V$ is coordinate vector with respect to $\displaystyle V$.
And $\displaystyle [T]_B = P^{-1}[T]_VP = P^{-1} AP$.

Now find eigenvalues $\displaystyle \bold{k}_1,\bold{k}_2,\bold{k}_3$ of $\displaystyle A$.
Then it would mean $\displaystyle P = [\bold{k}_1|\bold{k}_2|\bold{k}_3]$ would diagnolize $\displaystyle A$.

Now this means $\displaystyle [\bold{x}]_B = P[\bold{x}]_V \implies P^T[\bold{x}]_B = [\bold{x}]_V$

If you evaluate this equality at $\displaystyle \bold{b}_1$ you get $\displaystyle P^T \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = [\bold{b}_1]_V$

Thus, we get $\displaystyle \bold{c}_1 = \begin{bmatrix}d_1\\d_2\\d_3\end{bmatrix} = [\bold{b}_1]_V$ where $\displaystyle \bold{c}_1$ is first colomn of $\displaystyle P^T$. It follows that $\displaystyle \bold{b}_1 = d_1\bold{v}_1+d_2\bold{v}_2+d_3\bold{v}_3$ and that gives you what $\displaystyle \bold{b}_1$ is. Now do the same procedure for $\displaystyle \bold{b}_2,\bold{b}_3$.

3. I started to do what you said and computed the eigenvectors and im assuming it cant be right as the eigenvalues and vectors are probably the most hideous thing ive ever seen in my life ( done on maple). The answer that the book tells me is very simply and they seem to just get A out of no where.

edit: nm i got it thanks i was thinking stupidly.