Evaluating a cubic at a matrix only knowing only the eigenvalues

Question

I know the eigenvalues of a matrix, $A$, are $2,8,9$. How do I use this to find $m$ in the equation $$-4A^3+76A^2-424A+596I_3=mI_3$$ (where $I_3$ is the identity matrix for a $3\times 3$ matrix).

Answer 1

The content of the Cayley-Hamilton theorem is that every matrix satisfies its own characteristic equation.

Eigenvalues are, by definition, the roots of the characteristic polynomial $\det(x-A).$ If the eigenvalues of $A$ are $2, 8, 9$, then $(x-2)(x-8)(x-9)$ divides the characteristic polynomial of $A$, by the factor theorem. If it's a $3\times 3$ matrix, its charactersitic polynomial is of degree three, so then this is the characteristic polynomial.

Finally note that

$$(x-2)(x-8)(x-9)=x^3-19x^2+106x-144=-\frac{1}{4}(-4x^3+76x^2-424x+576).$$

So we have that

$$-4A^3+76A^2-424A+576=0$$

or $$-4A^3+76A^2-424A+596=20,$$

So $m=20.$

Answer 2

It doesn't matter if $A$ is $3\times3$ or larger. If $(\lambda,x)$ is an eigenpair of $A$, then $$ mx=(-4A^3+76A^2-424A+596I_3)x=(-4\lambda^3+76\lambda^2-424\lambda+596)x $$ and hence $m=-4\lambda^3+76\lambda^2-424\lambda+596$. Take $\lambda$ as any one of $2,8,9$, you obtain the same answer $m$.