Diagonalizable and finding Complex Matrix

Question

Consider the matrix

$\beta = \begin{pmatrix} 7 &3 &-4 \\ -2&-1 &2 \\ 6&2 &-3 \end{pmatrix}$ over the complex numbers.

Explain/show that it is Diagonalizable and find a complex matrix $\gamma$ and a diagonal matrix $\delta$ such that $\gamma ^{-1} \beta \gamma=\delta$

Effort so far

I have shown that it is indeed not possible over the real numbers and I know that to diagonalize a matrix $A$

Find the eigenvalues of $A$ using the characteristic polynomial.

For each eigenvalue $λ$ of $A$ , compute a basis $B λ$ for the $λ$-eigenspace.

If there are fewer than $n$ total vectors in all of the eigenspace bases $B λ$ , then the matrix is not diagonalizable. I am having trouble with showing this over the complex numbers and to find the matrices $\gamma , \delta$

Answer 1

The characteristic polynomial of $\beta$ is $-\lambda^3+3\lambda^2-\lambda+3=(3-\lambda)(\lambda^2+1)$, whose roots are $3$, $i$, and $-i$. If you search for eigenvectors corresponding to these eigenvalues, you will see that, for instance:

• $(1-i,2i,2)$ is an eigenvector corresponding to the eigenvalue $i$;
• $(1+i,-2i,2)$ is an eigenvector corresponding to the eigenvalue $-i$;
• $(1,0,1)$ is an eigenvalue corresponding to the eigenvalue $3$.

So, if you take$$M=\begin{bmatrix}1-i&1+i&1\\2i&-2i&0\\2&2&1\end{bmatrix}$$(the columns of $M$ are the eigenvectors that I got) then$$M^{-1}\beta M=\begin{bmatrix}i&0&0\\0&-i&0\\0&0&3\end{bmatrix}.$$