I have a matrix
$$
A = \begin{pmatrix}
3 & 1 & -1 \\
-1 & 1 & 1 \\
2 & 2 & 0
\end{pmatrix}.
$$
For this matrix, I have found eigenvalues of $2$ and $0$ and eigenvectors of
$$
e_2 = \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}
\qquad \text{and} \qquad
e_0 = \begin{pmatrix}
0 \\
1 \\
1
\end{pmatrix}.
$$
How do I find a matrix $T$ such that $T^{-1}AT$ is diagonal from these eigenvectors?
Best Answer
An $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. I see that you found one eigenvector corresponding to the eigenvalue $\lambda =2$. However, I suggest that the eigenspace of $A$ corresponding to the eigenvector $\lambda =2$ is two-dimensional.
Recall that one way to find the eigenvectors of the matrix $A$ corresponding to the eigenvalue $\lambda$ is to find the vectors $x$ which satisfy $(A- \lambda I)x=0$. Now $$(A- 2I)=\begin{pmatrix} 1&1&-1\\-1&-1&1\\2&2&-2 \end{pmatrix}.$$ This row-reduces to $$\begin{pmatrix} 1&1&-1\\0&0&0\\0&0&0 \end{pmatrix}.$$ Now the dimension of the eigenspace of $A$ corresponding to the eigenvalue $\lambda$ is equal to the dimension of the nullspace of $(A-\lambda I)$, which is $2$. So if $$x= \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix},$$ our system of equations that satisfies $(A-2I)x=0$ is $x_1+x_2-x_3 =0$. You found that one combination that satisfies that equation is $$x= \begin{pmatrix} 1\\0\\1 \end{pmatrix}.$$ Another one is $$x = \begin{pmatrix} -1\\1\\0 \end{pmatrix}.$$ These are your eigenvectors corresponding to the eigenvalue $\lambda =2$. Note that these vectors are linearly independent. Therefore they span a two-dimensional eigenspace.
Use these along with a correct eigenvalue for $\lambda =0$ to form the columns of your diagonalizing matrix. (Check your $e_0$).