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.
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: $a \neq 1$.