Finding the minimal polynomial when multiple eigenvalues have higher multiplicity than one

Question

I am learning about minimal polynomials with respect to determining the largest size of a Jordan block for a given eigenvalue. My question is:

when we have two or more eigenvalues with multiplicity of two or more, how can we be sure when we have found the minimal polynomial?

For example, let's say our characteristic polynomial is $(t-2)^3 (t+5)^2$. So we test the simplest polynomials first to see if the given matrix $A$ will yield a zero . . . We try first $(t-2) (t+5)$ and let's say it is not zero. So here's the crux of my question:

what if my next try is $(t-2)^2 (t+5)$ and I get a zero? Wouldn't I also have to try $(t-2) (t+5)^2$? What if both of these give me a zero?

Answer 1

The situation you described in your example can never happen because the minimal polynomial $m(x)$ of a linear transformation $T$ is the unique monic polynomial of lowest degree such that $m(T)$ = 0.

To prove this, we proceed by contradiction:

Assume $p, q$ be two distinct minimal polynomials (meaning they are monic and of equal degree and $p(T) = q(T) = 0$). Then let $r(x) =p(x)-q(x)$ which has degree strictly less than $p, q$ (because $p, q$ are monic). Then $r(T) = p(T) - q(T) = 0$, which is a contradiction, because the degree of $r$ is less than the degree of $p$ and $q$, so $r$ should be the minimal polynomial.

In your example, you can determine from the characteristic polynomial that the transformation has exactly two distinct eigenvalues, $2$ and $-5$, making the maximum size of the Jordan blocks corresponding to these eigenvalues at least 1, meaning both $(T-2)$ and $(T+5)$ must be factors of the minimal polynomial. If $(T-2)(T+5) \neq 0$, and $(T-2)^2(T+5) = 0$, then this must be the minimal polynomial because of uniqueness; you do not have to try the other option. It cannot be the case that $(T-2)(T+5) \neq 0$, $(T-2)^2(T+5) = 0$ and $(T-2)(T+5)^2 = 0$, as it would contradict uniqueness of the minimal polynomial.

Answer 2

What if my next try is (t-2)^2 (t+5) and I get a zero? Wouldn't I also have to try (t-2) (t+5)^2? What if both of these give me a zero?

For the first question, the answer is no. In fact, the set $\{P\in\Bbb K[x]:P(A)=0\}$ (where $A$ is the matrix and $\Bbb K$ a field containing its coefficients, like $\Bbb R$ or $\Bbb C$) contains all the multiples of a single polynomial, called the minimal polynomial $\mu_A$ of $A$. If you know that $(t-2)(t+5)$ doesn't vanish at $A$ whereas $(t-2)^2(t+5)$ does, you see that $\mu_A$ (whose roots are 2 and 5, the same as for the characteristic polynomial) is necessarily $(t-2)^2(t+5)$. So, using what I said before, we see that every polynomial vanishing at $A$ is a multiple of $\mu_A=(t-2)^2(t+5)$. Since $(t-2)(t+5)^2$ is not a multiple of it, it can't vanish at $A$ (otherwise, their gcd $(t-2)(t+5)$ would also vanish at $A$ and would be $\mu_A$).

When we have two or more eigenvalues with multiplicity of two or more, how can we be sure when we have found the minimal polynomial?

In general, if the characteristic polynomial writes as $$\chi_A(x)=(x-\lambda_1)^{k_1}\dots(x-\lambda_p)^{k_p}$$ then the minimal polynomial $\mu_A$ is of the form $$\mu_A(x)=(x-\lambda_1)^{\ell_1}\dots(x-\lambda_p)^{\ell_p}$$ where $\forall i\in\{1,\dots,p\}, 1\leq \ell_i\leq k_i$. Hence you have exactly $\displaystyle \prod_{i=1}^p k_i$ possibilities to check for $\mu_A$.

Another method for finding $\mu_A$ is by looking at the Jordan form of $A$ (which has to be computed before, of course!).

Rule: If there are several Jordan blocks corresponding to the eigenvalue $\lambda_i$, then $$k_i=\sum\text{(sizes of these blocks)}$$ and $$\ell_i=\max\text{(sizes of these blocks)}.$$