Finding The Minimal Polynomial Of Symmetric/Hermitian Matrix

Question

Given a specific symmetric/hermitian matrix, one can easily find the characteristic polynomial, can the minimal polynomial can be found without testing out lower powers polynomials with as a normal operator can be unitary diagonalized, and a matrix is diagonalized $\iff$ the minimal polynomial is a product of linear elements?

Answer 1

For symmetric $A$, the eigenvalues are real, which is trivial for the real field, and not difficult to show over the complex field. If $\lambda$ is real, then $$ \mathcal{N}((A-\lambda I)^2)=\mathcal{N}(A-\lambda I) $$ because, if $(A-\lambda I)^2x=0$, then

$$ 0=\langle (A-\lambda I)^2 x,x\rangle = \langle (A-\lambda I)x,(A-\lambda I)x\rangle = \|(A-\lambda I)x\|^2. $$ Thus, the minimal polynomial of a symmetric $A$ has no repeated factors. So the minimal polynomial is obtained from the characteristic polynomial by eliminating repeated factors.

Answer 2

Along the same lines as the other answer, suppose $A=A^\dagger$, $\lambda\in\mathbb C$, and $x$ is such that $(A-\lambda)^n x=0$ for some $n\in\mathbb N$.

Being $A$ Hermitian, this implies $\lambda\in\mathbb R$ (because it implies $Az=\lambda z=0$ for some normalised $z$, hence $\lambda=z^\dagger A z\in\mathbb R$).

If $n=2$, then $(A-\lambda)^2x=0$, which implies that $$\langle x,(A-\lambda)^2 x\rangle = \|(A-\lambda)x\|^2=0,$$ and thus $(A-\lambda)x=0$.

If $n>2$, define $y\equiv(A-\lambda)^{n-2}x$, so that $(A-\lambda)^2y=0$. By the argument above, we thus have $(A-\lambda)y=0$, and therefore $(A-\lambda)^{n-1}x=0$. Just apply the same reasoning until we get to $(A-\lambda)x=0$ and we are done.