How to find a 3x3 matrix

Question

How would you find a $3 \times 3$ matrix given that it scales everything on the plane $x + 2y + 3z = 0$ once, and scales $[0, −1, 1]$ twice.

Answer 1

Okay, so we have to do some diagonalization.

The vectors that make up the plane $x+2y+3z=0$, are the two vectors that are orthogonal to $\begin{bmatrix} 1\\2\\3 \end{bmatrix}$, which are $v_2=\begin{bmatrix} -2\\1\\0 \end{bmatrix}$ and $v_3=\begin{bmatrix} -3\\-6\\5 \end{bmatrix}$.

Now, we know that the eigenvectors $v_2$ and $v_3$ have eigenvalue $1$.

And the eigenvector $v_1 = \begin{bmatrix} 0\\-1\\1 \end{bmatrix}$

So our eigenvalue matrix is $D=\begin{bmatrix} 2&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$

We also have the eigenvector matrix which is $P=\begin{bmatrix} 0&-2&-3 \\ -1&1&-6 \\ 1&0&5 \end{bmatrix}$

Our transformation matrix will be $T=PDP^{-1}$.

Therefore, $T=\begin{bmatrix} 1&0&0 \\ -1&-1&-3 \\ 1&2&4 \end{bmatrix}$