Find $a$ in the following matrix

Question

I have the following question :

matrix $A$ isn't diagonalizable while $a \in R$

$$A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & a & a-2 \\ 0 & -2 & 0 \end{pmatrix}$$

Find $a$.

I don't really have any idea how to start, how to choose a "way" for this problem.

Any ideas?

Thank you!

Answer 1

In order for a matrix to not be diagonalizable, you need that the geometric multiplicity is less than the algebraic multiplicity. Meaning.. the number of (linearly independent) eigenvectors is less than the number of roots of the characteristic equation. Writing out the characteristic equation:

$$(3-\lambda)((a-\lambda)(-\lambda)-(a-2)(-2)) = (3-\lambda)(\lambda^2 - \lambda a + 2a-4).$$

The roots to this are $3$ (of course) and

$$\frac{a \pm \sqrt{a^2 - 4(2a-4)}}{2} = \frac{a \pm \sqrt{a^2 - 8a + 16}}{2} = \frac{a \pm (a-4)}{2}$$

which gives $2$ and $a-2$. For which value(s) of $a$ do we get a multiple root (meaning two or more eigenvalues are the same)? Does this/these choice(s) of $a$ give rise to a non-diagonalizable matrix?