Determine if matrix is diagonizable I am given matrix $A$:
\begin{pmatrix} 0 & 0 & -2\\1&2&1\\1&0&3\end{pmatrix}
I found the eigenvalues to be 1 and 2 by computing $\det(A-I\lambda)$.
Since the matrix is $3\times 3$ and there are only two distinct eigenvalues, I do not know what to do next. I know that there being n distinct eigenvalues proves that $A$ is diagonizable but the converse is not true so where do I go from here?
 A: A quick way to check is to see if $(A-I)(A-2I)=0$. If it is, then it is diagonalizable.
In general, for any square matrix $A,$ if the characteristic polynomial is $f(\lambda)=\det(A-\lambda I)$, then $$h(x)=\frac{f(x)}{\gcd(f(x),f'(x))}$$ has the same roots as $f$ but with no repeated roots, and $A$ is diagonalizable if and only if $h(A)=0$.
A: Check to see whether there are two linearly independent eigenvectors for the repeated eigenvalue.
If there are two, then the matrix can be diagonalised, if not (and there is only one vector for the eigenvec. of multiplicity two), then the matrix cannot.
A quick google of 'diagonalisability' should help you find a full lemma
A: The characteristic polynomial of 
$$
A=\left[\begin{array}{rrr}
0 & 0 & -2 \\
1 & 2 & 1 \\
1 & 0 & 3
\end{array}\right]
$$
is $\chi_A(t)=t^{3} - 5\,t^{2} + 8\,t - 4=(t - 1) \cdot (t - 2)^{2}$. Thus the eigenvalues of $A$ are $\lambda=1$ and $\lambda=2$. The algebraic multiplicities of the eigenvalues of $A$ are $\DeclareMathOperator{am}{am}\am_A(1)=1$ and $\am_A(2)=2$. 
Now, recall that the geometric multiplicity of an eigenvalue $\lambda$ of $A$ is defined as $\DeclareMathOperator{gm}{gm}\gm_A(\lambda)=\dim\DeclareMathOperator{Null}{Null}\Null(\lambda\cdot I-A)$. The inequality $1\leq \gm_A(\lambda)\leq\am_A(\lambda)$ is always respected. Thus the we have $\gm_A(1)=1$ and $\gm_A(2)=1$ or $\gm_A(2)=2$. Also recall that $A$ is diagonalizable if and only if each eigenvalue $\lambda$ of $A$ satisfies $\am_A(\lambda)=\gm_A(\lambda)$.
So, to check if $A$ is diagonalizable, we need only check if $\gm_A(2)=2$. To do so, note that
$$
2\cdot I-A=\left[\begin{array}{rrr}
2 & 0 & 2 \\
-1 & 0 & -1 \\
-1 & 0 & -1
\end{array}\right]
$$
One easily checks that $2\cdot I-A$ has rank one. The rank-nullity theorem then implies that $\gm_A(2)=2$. Hence $A$ is diagonalizable.
