# 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.

What the author is doing is using that if a matrix $$M$$ satisfies a polynomial $$p(t)$$, the minimal polynomial of $$M$$ divides $$p(t)$$. As all the eigenvalues of $$M$$ appear as roots of the minimal polynomial, you get that the eigenvalues of $$M$$ are contained in the set $$\{2,3\}$$.

Suppose that $$p(M)=0$$ for some square matrix $$M$$ and some polynomial $$p(\lambda)=\lambda^k+a_{k-1}\lambda^{k-1}+\cdots + a_{1}\lambda+a_0.$$ Then $$p(M)-p(\lambda)I = -p(\lambda)I.$$ You can rewrite the left side in order to obtain an inverse for $$M-\lambda I$$ for any $$\lambda$$ for which $$p(\lambda)\ne 0$$ as follows: $$(M-\lambda I)q(\lambda,M)=q(\lambda,M)(M-\lambda I)=-p(\lambda)I$$ Therefore $$M-\lambda I$$ is invertible if $$p(\lambda)\ne 0$$. So the only possible eigenvalues of $$M$$ are the solutions of $$p(\lambda)=0$$. That does not mean every root of $$p(\lambda)$$ is an eigenvalue because $$q(\lambda,M)=0$$ could occur. But it is certainly the case that every eigenvalue of $$M$$ is a root of $$p(\lambda)$$.

In your case, $$p(M)=0$$ where $$p(\lambda)=\lambda^2-5\lambda+6$$. So, the eigenvalues of $$M$$ must be roots of $$p$$, which are $$3$$ and $$2$$. That does not mean that both $$2$$ and $$3$$ are eigenvalues. But $$2$$ and $$3$$ are the only possible eigenvalues. Out of your possible answers, the only possible legitimate answer is (b) $$2,2,3$$ because neither $$1$$, nor $$5$$, nor $$6$$ are possible eigenvalues, as they are not roots of the annihilating polynomial $$p$$.

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)