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Find the values of the parameter $a\in\Bbb R$ for which the Jordan normal form of $$M=\begin{pmatrix}1&1&-1\\0&a-2&4\\0&-1&a+2\end{pmatrix}$$ contains a Jordan block of order $2$.

I calculated the characteristic polynomial, algebraic multiplicity and geometric multiplicity. However I obtain that geometric multiplicity is always one compared to algebraic multiplicity of 2, so I wanted to check.

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    $\begingroup$ So, what did you try? $\endgroup$ – Pedro Tamaroff Nov 3 '16 at 17:05
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    $\begingroup$ I calculated the characteristic polynomial, algebraic multiplicity and geometric multiplicity. However I obtain that geometric multiplicity is always one compared to algebraic multiplicity of 2, so I wanted to check. $\endgroup$ – oggyvukovich Nov 3 '16 at 17:08
  • $\begingroup$ @oggyvukovich good. Well, for future reference, note that you should provide context when asking a question $\endgroup$ – Omnomnomnom Nov 3 '16 at 17:14
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We obtain a characteristic polynomial of $$ (x-1)(x^2 - 2ax + a^2) = (x-1)(x-a)^2 $$ When $a \neq 1$, we note that the rank of $M - aI$ is $2$ (that is, $a$ has a geometric multiplicity of $1$), indicating that the Jordan form of $M$ has a block of size $2$ associated with $a$.

However, in the case that $a = 1$, we find that $M - I$ has a rank of $2$ (that is, $a = 1$ has a geometric multiplicity of $1$), indicating that the Jordan form of $M$ consists of one block of size 3.

So, the answer: $a \neq 1$.

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