# Possibilities of minimal polynomial for a matrix

I came across a problem recently in my linear algebra studies that went something like this:

Let $A$ be a linear transformation on a finite-dimensional space $V$ with characteristic polynomial $(x - 1)^6(x + 6)^2(x - 9)$ and minimal polynomial $m_A(x) = (x - 1)(x + 6)(x -9)$. Let $B$ be a linear transformation which commutes with $A$ and has characteristic polynomial $x^3(x - 2)^6$. Describe all possible minimal polynomials of $B$.

Now, via Cayley-Hamilton, I know that $m_B(x)$ must divide $x^3(x - 2)^6$, so $m_B(x) = x^a(x - 2)^b$, where the possibilities for $a$ are $1, 2, 3$ and for $b$ are $1, 2, 3, 4, 5, 6$, giving us a total of $18$ initial possibilities for the minimal polynomial. Moreover, I know that $A$ and $B$ are simultaneously triangulable, since they commute and operate on a finite-dimensional space $V$. Furthermore, I know $A$ is diagonalizable, since its minimal polynomial has only simple roots; and $A$ is invertible, since none of its eigenvalues is zero.

I would like to straightforwardly limit my choices for $m_B(x)$, but I cannot see a simple approach to eliminating any choices for $a$ and $b$. I tried to see what would happen if $a = b = 1$, but I got nowhere with this. Any ideas as to how I should limit my choices for $a$ and $b$?

## 1 Answer

So you know that $A$ has eigenspaces of dimensions $6,2,1$. The main constraint is that each of these subspaces is $B$-stable since $A$ and $B$ commute. This means that each of the eigenspaces for $A$ can be separately decomposed into Jordan blocks for $B$.

Now you know that the Jordan blocks for $B$ add up in size to $3$ for $\lambda=0$, and to $6$ for $\lambda=2$. This means for instance that if there is a single Jordan block of size$~3$ for $\lambda=0$, then it must be inside the eigenspace of dimension$~6$ for$~A$, and this only leaves space for Jordan blocks of size at most$~3$ for $\lambda=2$. So if in your terms $a=3$ then $1\leq b\leq3$. On the other hand if $a<3$, then its Jordan block can be lodged inside the $2$-dimensional eigenspace, and one only can say $1\leq b\leq6$ as you already knew.

• I like this approach. It appeals to the basic behaviors of Jordan blocks and simply works with relative sizes and what blocks can "fit." Thank you!! – d4rk_1nf1n1ty Apr 16 '15 at 5:54