Determining similar matrices I have this matrix
$$A= \begin{bmatrix}1 &0& 2\\0&-1&-2\\2&-2&0\end{bmatrix}$$
I found the eigenvalues to be $0, 3, -3$
I am tasked with finding if $A$ is similar to a diagonal matrix and find the $M$ such that
$M^{-1}AM $ is a diagonal matrix..
I have no idea how I would find M that meets the condition.
Thanks.
 A: Given a 3 by 3 matrix with 3 distinct eigenvalues, you know you will have 3 eigenvectors. Once you find these three eigenvectors, by finding $null(A-\lambda I)$ for $\lambda=3,-3,0$ in your case, giving you three linearily independent vectors. You glue these eigenvectors together to form a matrix $M$ with:
$D=M^{-1}AM$ where D is diagonal. So in your case, $M=[v_{3}v_{-3}v_{0}]$  where each v is an eigenvector. 
A: Any symmetric matrix ($A^{T} = A$) is diagonalizable (similar to a diagonal matrix). So this tells you that the answer is yes right away. 
Also, if you have an $n\times n$ matrix and you find that it has $n$ distinct eigenvalues, then it is diagonalizable. This is because diagonalizable is equivalent to there being a basis consisting of eigenvectors, and eigenvectors corresponding to different eigenvalues are linearly independent.
To find the matrix $M$, you just need to find a basis of eigenvectors. Since you have 3 distinct eigenvalues, you just need to find an eigenvector corresponding to each of them. Once you have these eigenvectors, make them the columns of the matrix $M$ and you are done. Note here that the order you put them in as columns doesn't matter, it will just effect what order the eigenvalues show up on the diagonal of the diagonal matrix.
