Diagonalizable and finding Complex Matrix

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 $$\gamma , \delta$$

• the way you wrote it, the columns of $\gamma$ are eigenvectors. Dec 30 '20 at 22:55
• this matrix has real spectrum wolframalpha.com/input/… Dec 30 '20 at 22:59
• @janmarqz No, it as not. The matrix that you have given to WolframAlpha is not $\beta$. Dec 30 '20 at 23:20
• @WillJagy what do you mean exactly?
– user831870
Dec 30 '20 at 23:28
• @bymathformath: this is because has three different eigenvalues and then three different eigenvectors giving the matrix that allows the diagonalization as indicated by the WolframAlpha result Dec 31 '20 at 2:16

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$$M^{-1}\beta M=\begin{bmatrix}i&0&0\\0&-i&0\\0&0&3\end{bmatrix}.$$

• I did not find these eigenvectors for the different eigenvalues. I found $i$ to give the vector $(1/2 + i/2, -i, 1)$ and $-i$ to give the vector $(1/2 - i/2, i, 1)$. And how would I argue for diagonalizeability?
– user831870
Dec 30 '20 at 23:33
• @bymathformath So what? That changes nothing. For each eigenvalue, there are infinitely many eigenvectors. Dec 30 '20 at 23:35
• @bymathformath Using other eigenvectors, you get another matrix $M$. But the diagonal matrix that you get at the end is always the same. Dec 30 '20 at 23:39
• Okay I see. why is that the case that this always works i.e any family of eigenvectors gives the same diagonal matrix in the end?
– user831870
Dec 30 '20 at 23:40
• @janmarqz I've edited my answer . Thank you. Dec 31 '20 at 7:55