Using the converse if Cayley Hamilton theorem My textbook says

: Let $M$ be a $3 \times 3$ Hermitian matrix which satisfies the matrix equation
$$
M^{2}-5 M+6 I=0
$$
Where $I$ refers to the identity matrix. Which of the following are possible eigenvalues of the matrix $M$
(a) (1,2,3)
(b) (2,2,3)
(c) (2,3,5)
(d) (5,5,6)

Then it proceeds as:

According to Cayley-Hamilton theorem, we can write $\lambda^{2}-5 \lambda+6=0 \Rightarrow \lambda=2,3$ Correct option is (b)

It's clear that the author has used the Cayley Hamilton theorem but in reverse but how can we use the converse Cayley Hamilton theorem? I've read that the converse of Cayley Hamilton theorem doesn't hold in general so what's the author doing here?
I'd be glad if someone pointed out my mistake. Much thanks.

The Cayley-Hamilton theorem has nothing to do with this business.
If $\lambda$ is an eigenvalue of $M$, then $Mv=\lambda v$, for some $v\ne0$. It follows that
$$
(M^2-5M+6I)v=(\lambda^2-5\lambda+6)v
$$
and since this is $0$ by assumption, we conclude that $\lambda^2-5\lambda+6=0$. Therefore $\lambda=2$ or $\lambda=3$.
After this you cannot conclude anything else from the data, because it could well be that $M$ has just the eigenvalue $2$ or just $3$: indeed $M=2I$ and $M=3I$ satisfy the given condition and are Hermitian. You can say nothing about the multiplicities of the eigenvalues, because also
$$
\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}
$$
satisfy the condition (and are Hermitian).
However, you can certainly exclude (a), (c) and (d)
