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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?

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    $\begingroup$ $$A-\lambda_1I=\left( \begin{array}{ccc} -1 & -1 & -2 \\ -1 & -1 & -2 \\ -2 & -2 & {\color{red}{-4}} \\ \end{array} \right)$$ $\endgroup$
    – Dario
    Commented Jun 29, 2014 at 12:26
  • $\begingroup$ Thanks. I kind of expected a stupid mistake like that. But simply didn't see it. $\endgroup$ Commented Jun 29, 2014 at 12:39

1 Answer 1

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All real valued symmetric matrices are diagonalizable. They are also called self adjoint.

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