# for which values of a is the matrix not diagonalizable

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.

• the matrix you typed in is symmetric, it is automatically diagonalizable. In fact, the eigenvalues are easy (you show them) and it is easy to make an orthogonal matrix $P$ with columns made of (normalized) eigenvectors – Will Jagy Apr 5 '20 at 18:45

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.
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.
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 $$a$$.