# Does the Cayley-Hamilton theorem go both ways

Say a $$n\times n$$ matrix $$A$$ is diagonalizable and has eigenvalues $${\lambda_1,\lambda_2,\dots,\lambda_n}$$. These eigenvalues satisfy: $$p(\lambda_i)=\det(A-\lambda_i I)=0.$$ From the Cayley-Hamilton theorem we know that: $$p(A)=0.$$ If we were given a matrix polynomial initially, would we be able to conclusively say anything about the eigenvalues of the matrix? For example, if we knew that a $$3\times 3$$ matrix $$A$$ satisfied: $$A^3-A=0$$ if we did the Cayley-Hamilton theorem backwards, we could potentially say that $$A$$ matrix has eigenvalues of $$0,1,-1$$. Is this true for all $$A$$ that satisfy the equation? Does the theorem work backwards?

• Certainly, if $A$ satisfied the equation $A^3 - A = 0$, then $A$ could possibly be the zero matrix, which has neither $1$ nor $-1$ as an eigenvalue.
– kobe
Dec 22, 2021 at 19:50
• Dec 22, 2021 at 21:31

If $$p$$ is a polynomial and $$p(A) = 0$$, then it can be deduced that the eigenvalues of $$A$$ are roots of $$p$$. (As noted in other answers and comments, it does not necessarily follow that every root of $$p$$ is an eigenvalue of $$A$$.)
Proof: Let $$\lambda$$ be an eigenvalue of $$A$$. By definition, there is a nonzero vector $$v$$ with $$Av = \lambda v$$. From $$Av = \lambda v$$ it follows that $$A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda^2 v$$, that $$A^3 v = A(A^2 v) = A(\lambda^2 v) = \lambda^2 Av = \lambda^3 v$$, and more generally $$A^n v = \lambda^n v$$ for any positive integer $$n$$, which implies that $$q(A) v = q(\lambda) v$$ for every polynomial $$q$$. Taking $$q = p$$ to be the polynomial for which $$p(A) = 0$$, we deduce that $$0 v = p(\lambda) v$$, which implies that $$p(\lambda) v = 0$$, which (since $$v$$ is a nonzero vector) implies that $$p(\lambda) = 0$$, i.e., that $$\lambda$$ is a root of $$p$$.
Note that this argument requires only the definition of "eigenvector" and not the theory of polynomials. In particular, we do not need to know that there exists a "minimal" polynomial of $$A$$, or identify how other polynomials relate to it. But that broader theory is certainly crucial to a general understanding of operators and the polynomial relations that they satisfy.
• So the answer is "almost" yes. In the OP example, we have to change a word: "matrix $A$ has eigenvalues among $\{0,1,−1\}$" Dec 22, 2021 at 19:58