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For which values of $a$ and $b$ is the matrix $$ \begin{pmatrix} 0 & a\\ b & 0 \end{pmatrix} $$ diagonalizable over $\mathbb{C}$?

I know that if $a = -1$ and $b = 1$, then the matrix is diagonalizable. However, I am not sure if that is the only solution or how to go about finding solutions in general. Please help.

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Do you know how to find the eigenvalues? – Brian Fitzpatrick Apr 26 '13 at 2:57
Yes, I do--vaguely. What I tried doing was looking at the characteristic polynomial to help, and I came up with a difference of two squares. I don't really think that is correct though since it seems to easy. – Trixie Apr 26 '13 at 3:04

$$A=\begin{pmatrix}0&a\\b&0\end{pmatrix}\implies p_A(x):=\det(xI-A)=\begin{vmatrix}\;x&\!\!-a\\\!\!-b&\;x\end{vmatrix}=x^2-ab$$


$$p_A(x)=0\iff x=\pm\sqrt{ab}\implies$$

$$(1)\;\;\;\;\;\;\;\;\;\;ab\neq0\implies\;\text{there are two different eigenvalues and the matrix is diagonalizable}$$

$$(2)\;\;\;\;\;\;\;\;\;\;ab=0\;,\;a\neq 0\,\,\vee\,\,b\neq0\implies\text{ this is a non-zero nilpotent matrix and thus non-diagonalizable}$$

$$(3)\;\;\;\;\;\;\;\;\;a=b=0\implies\text{ the matrix's already diagonal...}$$

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That, again, was an act of god, @Amzoti...:P) Thanks for catching that and corrected. – DonAntonio Apr 26 '13 at 3:20
you are very welcome! Nice solution! +1 – Amzoti Apr 26 '13 at 3:21
That @julien , what to do!, was again the hand of god. It should have been non-diagonalizable. Thanks. – DonAntonio Apr 26 '13 at 16:26

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