Diagonalizing a 3x3 matrix Prove that matrix $A$ is diagonalizable, find the bases for the eigenspaces, the diagonalizing matrix $P$, and compute $P^{-1} A P$ where 
$A= \left(\begin{array}{ccc} 2 & 0 & 3 \\
      0 & 3 & 0 \\
      0 & 0 & 3  \end{array}\right).$

The characteristic polynomial is $ \text{det}(A−λI)=(2−λ)(3−λ)^2$ so the eigenvalues of your matrix are $2$ and $ 3$. Therefore $2$ is an eigenvalue with algebraic multiplicity $1,$ and $3$ is an eigenvalue with algebraic multiplicity $2$. Recall the geometric multiplicity can also be described as the dimension of the nullspace of $A-λI$. 
So for $λ = 2$ we have $A-λI = \left(\begin{array}{ccc} 
      0 & 0 & 3 \\
      0 & 1 & 0 \\
      0 & 0 & 1  \end{array}\right)$ so the null space has dimension $1.$ Thus the geometric multiplicity for $λ = 2$ is $1$.
Now for $λ = 3$ we have $A-λI = \left(\begin{array}{ccc} 
      -1 & 0 & 3 \\
      0 & 0 & 0 \\
      0 & 0 & 0  \end{array}\right)$ so the null space has dimension $2.$ Thus the geometric multiplicity for $λ = 3$ is $2$.
Because the algebraic multiplicities equal the geometric multiplicities for all the eigenvalues, by definition the matrix is non-defective and hence diagonalizable.
For a good explanation of diagonalizability see Chapter 24 of "Numerical Linear Algebra" by Trefethen and Bau.
