Roots of minimal and characteristic polynomials

Solution 1:

First, show the following easy claim: Let $f(x)\in\mathbb{C}[x]$ be any polynomial. If $\lambda$ is an eigenvalue of $A$ associated with the eigenvector $v$, then $f(\lambda)$ is an eigenvalue of $f(A)$ associated with the eigenvector $v$. (prove by calculating $f(A)v$, using distributivity of multiplication over addition).
Now let $\lambda$ be a root of $\chi_A(t)$. Thus $\lambda$ is an eigenvalue and so has an associated eigenvector $v\neq 0$. Using the claim, we get $0=\mu_A(A)\cdot v=\mu_A(\lambda)\cdot v$ (since $\mu_A(A)=0$). Since $v\neq 0$, we get $\mu_A(\lambda)=0$.

Solution 2:

I remember of something like this : if $\mu_A(t)=\prod_{\lambda \in Sp(A)}(t-\lambda)$, then $\chi_A(t)=\prod_{\lambda \in Sp(A)}(t-\lambda)^{\nu_\lambda}$. The two polynomials have the same roots, the eigenvalues of $A$, and the order of multiplicity of $\lambda$ is $1$ in the minimal polynomial and $\nu_\lambda$ in the characteristic polynomial.

This is proved by Dennis Gulko's answer.

Further, the order of multiplicity $\nu_\lambda$ is the the dimension of the eigenspace $\nu_\lambda = \dim E_\lambda=\dim \ker (\lambda I - A)=\dim \{x|Ax=\lambda x\}$.

This is shown by looking at the restriction of $A$ to $E_\lambda$, where it acts as the homothety $x \rightarrow \lambda x$ and its matrix is the square diagonal matrix $\left( \matrix{\lambda&0&...&0 \\ 0&\lambda&...&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&...&\lambda} \right)$.

Solution 3:

Dennis Gulko has a good answer for the direction: every eigenvalue is a root of the minimal (and indeed of any annihilator) polynomial. (Summary: if a linear operator $A$ act on (an eigenvector) $v\neq0$ as scalar multiplication by$~\lambda$, then any polynomial $P[A]$ of $A$ acts on $v$ as scalar multiplication by $P[\lambda]$, so $P[A]=0$ implies $P[\lambda]=0$.)

With this answer I just like to note that you don't need the Cayley-Hamilton theorem* for the opposite direction: every root of the minimal polynomial is an eigenvalue. If one factors $\mu_A=(X-\lambda)Q$, then by minimality of $\mu_A$ the quotient$~Q$ cannot be an annihilator polynomial; then any nonzero vector in the image of $Q[A]$ is in the kernel of $(X-\lambda)[A]=A-\lambda I$, and therefore an eigenvector of $A$ for the (apparently) eigenvalue$~\lambda$.

*One may observe that "you don't need the Cayley-Hamilton theorem" is a recurring theme in my answers. Of course anyone is absolutely free to use the Cayley-Hamilton theorem anyway (especially if they can remember one of its proofs), unless it is for results leading up to a proof of said theorem.