Find a matrix $P$ that diagonalizes

Question

$$ A= \begin{bmatrix} 3 & -2 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 5 \ \end{bmatrix} $$ I can find the eigenvalues is $\lambda = 1$ and $\lambda = 5$ but when I apply them into $A$, I can't find the eigenvectors, can you help me with this part ?

Answer 1

for the eigenvalue $\lambda=1$:
$A(x,y,z)^t=(x,y,z)^t$ is equivalent to $$3x-2y=x$$ $$-2x+3y=y$$ $$5z=z$$ which defines the subspace $Sp\{(1,1,0)\}$. So an eigenvector for $\lambda=1$ is $(1,1,0)$. can you continue from here?

Answer 2

If you solve the system $A.\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)=5\begin{smallmatrix}x\\y\\z\end{smallmatrix}$, you will get a $2$-dimensional space. An obvious element of that space is $\left(\begin{smallmatrix}0\\0\\1\end{smallmatrix}\right)$ and another (less obvious) one is $\left(\begin{smallmatrix}-1\\1\\0\end{smallmatrix}\right)$. These two are linearly independent. Finally, $\left(\begin{smallmatrix}1\\1\\0\end{smallmatrix}\right)$ is an eigenvector of $A$ with eigenvalue $1$. So, if you define$$P=\begin{pmatrix}1&-1&0\\1&1&0\\0&0&1\end{pmatrix}$$(the columns are the eigenvectors), then$$P^{-1}.A.P=\begin{pmatrix}1&0&0\\0&5&0\\0&0&5\end{pmatrix}.$$