Find the eigenvalues and eigenvectors of the matrix.
$$A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & -1\\ 0 & -1 & 1 \end{bmatrix}$$
As we have seen in the lectures, these eigenvectors form an orthogonal basis with respect to the standard inner product $\mathbb{C}^3$ . By considering a basis transformation to an orthonormal basis of eigenvectors find a diagonalizing matrix $P$, and hence $B = P^{-1}AP$ where $B$ is diagonal. (Hint: $P^{-1} = P^{T}$ for an orthonormal basis to another.)
I've only got 1 eigenvalue to be $\lambda = 1, -1, 2$ with their corresponding eigenvectors. I am not sure where to go from here. Any help would be appreciated!! Thank you.
To find the eigenvalues you can use the characteristic polynom : $$det \left( \begin{bmatrix} 1-X & 1 & 0 \\ 1 & 0-X & -1\\ 0 & -1 & 1-X \end{bmatrix}\right)=det \left( \begin{bmatrix} 1-X & 0 & 1-X \\ 1 & -X & -1\\ 0 & -1 & 1-X \end{bmatrix}\right)=(1-X)det \left( \begin{bmatrix} 1 & 0 & 1 \\ 1 & -X & -1\\ 0 & -1 & 1-X \end{bmatrix}\right)=(1-X)det \left( \begin{bmatrix} 1 & 0 & 0 \\ 1 & -X & -2\\ 0 & -1 & 1-X \end{bmatrix}\right)=(1-X)(-X(1-X)-2)=(1-X)(X^2-X+2)=-(X-1)(X+1)(X-2)$$
So the eigen values are $-1,1,2$.
Can you find the eigen vectors from there ?
For $-1$. Let $x \in \mathbb{C}^3$ such as : $$Ax=-x\iff A\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} -x_1 \\ -x_2 \\ -x_3 \end{bmatrix} \iff \begin{bmatrix} x_1+x_2 \\ x_1-x_3 \\ -x_2+x_3 \end{bmatrix}=\begin{bmatrix} -x_1 \\ -x_2 \\ -x_3 \end{bmatrix} $$
So $x_2=-2x_1$ and $x_3=-x_1$. So the eigenvector associated to $-1$ is $\begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}$.
Do the same for the two other eigenvalues($1$ then $2$), concatenate the vectore you obtain in the order of obtention (first column is the vector associated with $-1$) and you obtain $P$ such as $A=PDP^{-1}$, where $D=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 2 \end{bmatrix}$.