Diagonalize the matrix $\begin{bmatrix}0&-4&-6\\-1&0&-3\\1&2&5\end{bmatrix}$ if possible
So I know that I can check to see if this is diagonalizable by doing $A = PDP^{-1}$ but I think I'm doing something wrong when I'm finding the eigenvectors.
The eigenvalues are $\lambda_1 = 1, \lambda_2 = 2$
I did $\det(A-\lambda_1 I)\vec{x} = \vec{0}$ and got $\begin{bmatrix}1&0&6\\0&1&0\\0&0&0\end{bmatrix}$ which gives me:
$x_1 +6x_3 = 0$
$x_2 = 0$
So I have 2 pivots so I know I will have $\mathbb{R}^2$ columns right?
So I get the eigenvector: $\begin{bmatrix}-6&0\\1&0\\0&1\end{bmatrix}$
Then I get for $\det(A - \lambda_2I) = \begin{bmatrix}1&2&3\\0&1&3/2\\0&0&0\end{bmatrix}$ so I get:
$x_1 + 2x_2 + 3x_2$
$x_2 = -3/2$
so the eigenvector $\begin{bmatrix}-2&-3\\1&-3/2\\0&1\end{bmatrix}$ is this correct or have I messed up on my eigenvectors?
UPDATE:
I'm also really confused because would $D = \begin{bmatrix}1&0\\0&2\end{bmatrix}$ and
$P = \begin{bmatrix}-6&0&-2&-3\\1&0&1&-3/2\\0&1&0&1\end{bmatrix}$ ?
To find the eigenvalues of $A$, we want to find $\lambda$ such that $A - \lambda I$ has a nontrivial null space (equivalent to the determinant being 0).
$$A-\lambda I = \begin{bmatrix} -\lambda & -4 & -6 \\ -1 & -\lambda & -3 \\ 1 & 2 & 5-\lambda \end{bmatrix}$$.
By either row reduction or taking the determinant, we can compute that the eigenvalues are indeed $1$ and $2$. We can find the eigenspaces and thus all of the eigenvectors by computing $Null(A-I)$ and $Null(A-2I)$.
For $\lambda = 1$:
$$A- I = \begin{bmatrix} -1 & -4 & -6 \\ -1 & -1 & -3 \\ 1 & 2 & 4 \end{bmatrix}$$.
After row reduction we get: $$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$ which has the same null space as $A-I$.
We then compute the null space by setting up the following system of equations: The eigenvectors will be $<x_1, x_2, x_3>$ such that the following equations hold:
$x_1 + 2x_3 = 0$
$x_2 + x_3 = 0$
Which is equivalent to...
$x_1 = -2x_3$
$x_2 = -x_3$.
So, taking $x_3 = 1$ (or you could choose any other nonzero value), $E(A, 1) = Null(A-I) = span(<-2, -1, 1>)$
Now, you need to do the same thing for $\lambda = 2$. i.e. Find the null space of $A-2I$, and you will have found all of the eigenvectors with eigenvalue $2$. This time, you will find that the null space is of dimension 2. As you will have 3 linearly independent eigenvectors total, $\mathbb{R}^3$ has a basis of eigenvectors of $A$, and thus, $A$ is diagonalizable. From there, you can use this basis of eigenvectors to compute $D$, and use the change of basis formula to compute $P$ and $P^{-1}$.