Given the matrix $$A=\left( \begin{array}{ccc} 0 & -1 & -2 \\ -1 & 0 & -2 \\ -2 & -2 & -3 \\ \end{array} \right)$$
It has the following characteristic polynomial: $-(\lambda-1)^2(\lambda+5)$, yielding eigenvalues $\lambda_1=1,\lambda_2=2$. Now the algebraic multiplicity for $\lambda_1$ is 2.
For the nullspace of $A-\lambda_1I$ I row reduce the matrix and get:
$$A-\lambda_1I=\left( \begin{array}{ccc} -1 & -1 & -2 \\ -1 & -1 & -2 \\ -2 & -2 & -3 \\ \end{array} \right) \Leftrightarrow \left( \begin{array}{ccc} 0 & 0 & 0 \\ -1 & -1 & -2 \\ 0 & 0 & 1 \\ \end{array} \right)$$
Therefore $\ker(A-\lambda_1I)=\langle \left( \begin{array}{ccc} -1 & 1 & 0 \\ \end{array} \right) \rangle$, so the geometric multiplicity is only 1, and a matrix is diagonalizable iff the algebraic multiplicity equals the geometric multiplicity. Mathematica tells me the matrix is diagonalizable, but I can only come up with two eigenvectors. Where is my mistake?
All real valued symmetric matrices are diagonalizable. They are also called self adjoint.