My professor gave us a question and I would like some help in it. The problem goes as follows: Let $T:\mathbb{R}^3\to\mathbb{R}^3$ be the linear transformation represented by: $T(x_1,x_2,x_3)=(x_1-x_2-x_3,x_1+3x_2+x_3,-3x_1+x_2-x_3)$
a) Find the standard matrix $A$ for T.
b) Show that $A$ is diagonalizable.
c) Compute $A^3$.
d) Find a basis $B$ for $\mathbb{R}^3$ such that the matrix $A$ relative to $B$ is diagonal.
Here is what I know:
a) $A=\begin{bmatrix} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -3 & 1 & -1 \end{bmatrix}$
b) $\det(\lambda I-A)=0$
where the characteristic polynomial is $\lambda^3-3\lambda^2-4\lambda+12$. Therefore the eigenvalues are $2,-2,3$. Since there are 3 distinct eigenvalues for the matrix A which is in $\mathbb{R}^3$, A is diagonalizable. However, I proceeded to find the bases because I need them for part c). Here's where I'm stuck. I found the basis for when $\lambda=2$ and got $p_1=\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$ and $p_2=\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$. However, the basis for $\lambda=2$ is $p_1=\begin{bmatrix} -1\\ 1\\ 1 \end{bmatrix}$.
Where did I go wrong? Edit: I included further calculations For $\lambda=2$,
$\begin{bmatrix} 1 & 1 & 1\\ -1 & -1 & -1\\ 3 & -1 & 3 \end{bmatrix}$ $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$=$\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
This reduces to $\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$=$\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
So, $x_1+x_3=0, x_2=0, x_3=0$. Then, $x_1=-t, x_2=s,x_3=t$
Hence, $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=$ $\begin{bmatrix} -t\\ s\\ t \end{bmatrix}$=$t\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}+$$s\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$
So, $p_1=\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$,$p_2=\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$
To find the eigenvectors for the basis, you have to compute the basis for $\ker(2I - A), \ker(-2I - A), \ker(3I - A)$. For example, I'll set the first.
$$\left[\begin{array}{ccc} -1 & -1 & -1 \\ 1 & 1 & 1 \\ -3 & 1 & -3 \end{array}\right] \left [\begin{array}{c} x \\ y \\ z \end{array}\right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] $$
Can you go on?