<s>[INST] How do I determine if matrix A is diagonalizable? I am trying to figure out how to determine the diagonalizability of the following matrix:
$A=\begin{pmatrix}
1 &0  &0  &0 \\ 
 2&1  & -3 & -2\\ 
 3& 0 & 0 &-9 \\ 
 -1&  0&  -1& 0
\end{pmatrix}$
There are two distinct eigenvalues: $λ_1=λ_4=1$,$λ_2=3$ and $λ_3=-3$.
Can someone help to define geometric multiplicity?
And I found that matrix is diagonalizable geometric multiplicity is equal to the algebraic multiplicity.
 [/INST] The characteristic polynomial is
$$
\det
\begin{pmatrix}
1-X &0  &0  &0 \\ 
 2&1-X  & -3 & -2\\ 
 3& 0 & 0-X &-9 \\ 
 -1&  0&  -1& 0-X
\end{pmatrix}
=(1-X)^2(3-X)(-3-X)
$$
You want to determine the geometric multiplicity of the eigenvalue $1$, that is the dimension of the eigenspace, which is the null space of $A-I$. Let's do an elimination:
\begin{align}
A-I=\begin{pmatrix}
0 &0  &0  &0 \\ 
 2&0  & -3 & -2\\ 
 3& 0 & -1 &-9 \\ 
 -1&  0&  -1& -1
\end{pmatrix}
&\to
\begin{pmatrix}
 -1&  0&  -1& -1 \\
 2&0  & -3 & -2\\ 
 3& 0 & -1 &-9 \\
0 &0  &0  &0
\end{pmatrix}
\\[6px]&\to
\begin{pmatrix}
 1&  0&  1& 1 \\
 0& 0  & -5 & -4\\ 
 0& 0 & -4 &-12 \\
0 &0  &0  &0
\end{pmatrix}
\\[6px]&\to
\begin{pmatrix}
 1&  0&  1& 1 \\
 0& 0  & 1 & 4/5\\ 
 0& 0 & 0 &-44/5 \\
0 &0  &0  &0
\end{pmatrix}
\\[6px]&\to
\begin{pmatrix}
 1&  0&  1& 1 \\
 0& 0  & 1 & 4/5\\ 
 0& 0 & 0 &1 \\
0 &0  &0  &0
\end{pmatrix}
\end{align}
So the matrix $A-I$ has rank $3$ and its null space has dimension $1$.
The matrix is not diagonalizable.
</s>