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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 $\gamma , \delta$

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  • $\begingroup$ the way you wrote it, the columns of $\gamma$ are eigenvectors. $\endgroup$
    – Will Jagy
    Dec 30 '20 at 22:55
  • $\begingroup$ this matrix has real spectrum wolframalpha.com/input/… $\endgroup$
    – janmarqz
    Dec 30 '20 at 22:59
  • $\begingroup$ @janmarqz No, it as not. The matrix that you have given to WolframAlpha is not $\beta$. $\endgroup$ Dec 30 '20 at 23:20
  • $\begingroup$ @WillJagy what do you mean exactly? $\endgroup$
    – user831870
    Dec 30 '20 at 23:28
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    $\begingroup$ @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 $\endgroup$
    – janmarqz
    Dec 31 '20 at 2:16
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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$$M^{-1}\beta M=\begin{bmatrix}i&0&0\\0&-i&0\\0&0&3\end{bmatrix}.$$

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  • $\begingroup$ 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? $\endgroup$
    – user831870
    Dec 30 '20 at 23:33
  • $\begingroup$ @bymathformath So what? That changes nothing. For each eigenvalue, there are infinitely many eigenvectors. $\endgroup$ Dec 30 '20 at 23:35
  • $\begingroup$ @bymathformath Using other eigenvectors, you get another matrix $M$. But the diagonal matrix that you get at the end is always the same. $\endgroup$ Dec 30 '20 at 23:39
  • $\begingroup$ 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? $\endgroup$
    – user831870
    Dec 30 '20 at 23:40
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    $\begingroup$ @janmarqz I've edited my answer . Thank you. $\endgroup$ Dec 31 '20 at 7:55

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