Inequality about the degree of minimal polynomial For finite-dimensional vector space $V$, there exist linear operators $A$ and $B$ on $V$ such that $AB=BA$ commutative relation holds.
If we define the $A$'s minimal polynomial degree by $\deg(A)$, how can I prove the inequality $\deg(A+B)\leq \deg(A)\deg(B)$?
I grasp the idea that in the minimal polynomial of $A+B$, I could expand the $(A+B)^k$ terms by exchanging
multiplication order of $A$ and $B$ but I can't proceed further.
 A: Note that $\deg(A)$ is the dimension of the subspace consisting of all polynomials of $A$.
Let $m = \deg(A), n = \deg(B)$. Every polynomial of $p(A)$ can be written as a linear combination of the powers $I,A,\dots,A^{m-1}$ of $A$, and every polynomial $p(B)$ can be written as a linear combination of the powers $I,B,\dots,B^{n-1}$ of $B$. We conclude that for any bivariate polynomial $p(x,y)$, $p(A,B)$ can be written as a linear combination of the elements of $S = \{A^jB^k : 0 \leq j \leq m-1, \ 0 \leq k \leq n-1\}$. Note that $S$ contains $\deg(A)\deg(B)$ elements, so its span is at most $\deg(A) \deg(B)$ dimensional.
Because
$
\operatorname{span}\{I,(A+B),(A+B)^2,\dots\} \subset \operatorname{span}(S),
$
we can conclude that
$$
\deg(A + B) = \dim \operatorname{span}\{I,(A+B),(A+B)^2,\dots\} \leq \dim \operatorname{span}(S) \leq \deg(A)\deg(B).$$

For another perspective, we could note that the map from $\Bbb F[x,y]$ to $p(x,y)$ defined by $p\mapsto p(A,B)$ is an $\Bbb F$-algebra homomorphism.
A: $Om(nom)^3$'s elegant answer really shows the key: $\deg(A)$ is the dimension of the subspace of all polynomials in $A$. I cannot do better. However, if you want to use binomial expansion to solve the problem, here is one way.
We may assume that the underlying field is algebraically closed, because minimal polynomial is invariant under field extension. Factor the minimal polynomial of $A$ into $\prod_{i=1}^k(x-\lambda_i)^{m_i}$, where the $\lambda_i$s are distinct. Let $V_i=\ker((A-\lambda_iI)^{m_i})$. Then $V=V_1\oplus V_2\oplus\cdots\oplus V_k$ and $(A-\lambda_iI)^{m_i}BV_i=B(A-\lambda_iI)^{m_i}V_i=0$. Hence $BV_i\subseteq V_i$. In other words, each $V_i$ is an invariant subspace of both $A$ and $B$.
We can prove that
$$
\deg(A+B)\le \deg(A) \deg(B).\tag{0}
$$
by showing that each the following lines is true:
\begin{align}
\deg(A+B)
&\le \sum_i \deg((A+B)|_{V_i})\tag{1}\\
&\le \sum_i \deg(A|_{V_i}) \deg(B|_{V_i})\tag{2}\\
&\le \sum_i \deg(A|_{V_i}) \deg(B)\tag{3}\\
&= \deg(A) \deg(B)\tag{4}\\
\end{align}
$(1),(3)$ and $(4)$ are clearly true. To prove $(2)$, we may show that $\deg((A+B)|_{V_i})\le \deg(A|_{V_i}) \deg(B|_{V_i})$ for each $i$. In other words, to prove $(0)$, it suffices to consider the special case where all eigenvalues of $A$ are equal to some $\lambda$. As $\deg(M)=\deg(M-\lambda I)$, we may further assume that $A$ is nilpotent. But then, if we interchange the roles of $A$ and $B$ and go through some similar arguments to the above, we see that we may also assume that $B$ is nilpotent.
Thus it suffices to prove $(0)$ for two commuting nilpotent matrices $A$ and $B$. Let $r,s\,(\ge1)$ be the indices of nilpotence of $A$ and $B$ respectively. Then $(0)$ is equivalent to the statement that $(A+B)^{rs}=0$. But this is true because $(r-1)+(s-1)\le rs$, i.e. if $p,q\ge0$ and $p+q=rs$, we must have $p\ge r$ or $q\ge s$. Hence each term $A^pB^q$ in the binomial expansion of $(A+B)^{rs}$ is zero.
