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Suppose we have matrix of the form
$$ A= \begin{bmatrix} a & -1 \\ 0 & a \\ \end{bmatrix} $$
and we would like to analyze its diagonalizability.
By taking the characteristic equation we quickly find $a^2 = 0$. In this case, can we say that this matrix has eigenvalue of 0 with infinite multiplicity?
You're computing the determinant of $A$, which is not the characteristic polynomial. The characteristic polynomial is rather $$ \det(A-XI)=\det\begin{bmatrix}a-X&-1\\0&a-X\end{bmatrix}=(a-X)^2 $$ which has one root (equal to $a$) with multiplicity $2$.
The matrix is not diagonalizable, because $$ A-aI=\begin{bmatrix}0&-1\\0&0\end{bmatrix} $$ has rank $1$.
Nope. This is a polynomial equation in $a$ of degree $2$, so that if there is only one $a\in\mathbb R$ with $a^2=0$ (and this is the case: any $a\neq 0$ will have $a^2>0$), then this $a$ will be a double root, that is, a root of multiplicity $2$.
In the end, you can never have an eigenvalue of multiplicity greater than $n$ for an $n\times n$ matrix, because the characteristic equation of an $n\times n$ matrix is always a polynomial equation of degree $n$.
