for which values of a is the matrix not diagonalizable

Question

For which values of a is the below matrix NOT diagonalizable:

\begin{pmatrix}-5&8&0\\ \:\:\:8&-5&0\\ \:\:\:0&0&a\end{pmatrix}

Take the determinant of: \begin{pmatrix}-5-λ&8&0\\ 8&-5-λ&0\\ 0&0&a-λ\end{pmatrix} to find the characteristic equation.

= (-5-λ)(-5-λ)(a-λ) - (8)(8)(a-λ)

= λ^2a + 10λa - 39a - λ^3 - 10λ^2 + 39λ

Find eigenvalues by setting above to zero

λ = a,3,-13

The matrix is diagonalizable if a is either 0, 3, or -13.

I'm not sure if there is any property I'm forgetting that is required to solve the problem.

Any guidance would be greatly appreciated.

Answer 1

As mentioned in the comment, all symmetric matrices are diagonalizable. Hence just conclude that no such value of $a$ exists and you are done.

Now, let's revisit what you did. You found the eigenvalues of the matrix, which are $a, 3$ and $-13$ by first expanding the characteristic polynomial. If we visualize the matrix as a block diagonal matrix, then we can see that $a$ is an eigenvalue and the remaining eigenvalues are the eigenvalues of the first $2 \times 2$ block.

Notice that we have the result that if all eigenvalues are distinct, then the matrix is diagonalizable.

In the event that the matrix is not symmetric and has some eigenvalue with algebraic multiplicity more than $1$, then we check if the geometric multiplicity is equal to the algebraic multiplicity.

Answer 2

As noted in a comment, the matrix is real symmetric, hence diagonalizable regardless of the value of $a$.

If you didn’t happen to know that, observe that $(0,0,1)^T$ is an eigenvector of $A$ with eigenvalue $a$. The other eigenvalues are those of the upper-left $2\times2$ submatrix, which you’ve computed to be $3$ and $-13$. If $a$ is not equal to either of these, then you have three distinct eigenvalues and the matrix is diagonalizable. On the other hand, if $a=3$ or $a=-13$, then the matrix is diagonalizable iff you can find two linearly-independent eigenvectors for that eigenvalue. That’s always possible, since the eigenvector that comes from the upper-left block must be of the form $(x,y,0)^T$, which is clearly linearly independent from $(0,0,1)^T$. Thus, the matrix is diagonalizable for all values of $a$.