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For which values ​​of $a$ matrix $A$ is ​​diagonalizable? $$A = \pmatrix{0&i\\i&a}$$

in the case that it is not diagonalizable determine a base of Jordan

Attempt: The minimal polynomial already factored in is: $p_A(x)=(x-(\frac{a-\sqrt{a^2-4}}{2}))(x-(\frac{a+\sqrt{a^2-4}}{2}))$.

If they are the two distinct roots then $A$ is diagonalizable because its minimum polynomial is equal to the characteristic one and we see that it has different linear factors. On the other hand the roots of the polynomial coincide when $a= \pm 2$. It is verified that the minimum polynomial is not $(x \pm 1)$, because it does not annul $A$ (when $a= \pm 2$). In this case $A$ is not diagonalizable and we will find the base of Jordan.Is it correct here?

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    $\begingroup$ It looks just fine and dandy to me. $\endgroup$ – DonAntonio Feb 10 at 20:38
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    $\begingroup$ For me it looks fine. $\endgroup$ – user376343 Feb 10 at 20:38
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    $\begingroup$ Yes, you are correct so far, as long as you don't write $\sqrt{a^2-4}$ when $a$ is a complex... Just note $\delta$ one square root of $a^2-4$, and proceed with this notation. $\endgroup$ – Nicolas FRANCOIS Feb 10 at 20:38

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