I looked around, and as far as I can tell, I haven't found this question anywhere else on SE, so if I somehow missed it, please pardon me.
I think this is probably a standard result, but I am having trouble proving it nonetheless. Here is the statement I wish to prove:
"Let $A,B \in M_{n \times n}(\mathbb{C})$ (i.e., the set of $n\times n$ matrices over $\mathbb{C}$). If for all $B$ such that $AB=BA$ one has $B \in \text{Span}\{I, A, A^2,\ldots, A^{n-1}\}$, then $A$ has $n$ distinct eigenvalues."
I have tried messing around with the minimal polynomial and the Jordan form. My instinct is to show that the given condition implies that the Jordan blocks are all of size $1 \times 1$ (hence my concern with the minimal polynomial), and that these blocks are distinct. However, I am clearly not seeing the crucial connection here.
Incidentally, I think I have proven the converse.
If you are willing to share any hints/thoughts/solutions, I would be very appreciative of your input.
the thing you are trying to prove is false. The correct condition is that every eigenvalue occurs in just one Jordan block.
Find out, by hand, what matrices $B$ commute with $$ A = \left( \begin{array}{cc} 7 & 1 \\ 0 & 7 \end{array} \right) $$ $$ B = \left( \begin{array}{cc} p & q \\ r & s \end{array} \right) $$ $$ AB = \left( \begin{array}{cc} 7p+r & 7q+s \\ 7r & 7s \end{array} \right) $$ $$ BA = \left( \begin{array}{cc} 7p & p+7q \\ 7r & r + 7s \end{array} \right) $$
As $7p+r = 7p,$ we must have $r=0.$ As $7q+s=p+7q,$ we must have $s=p.$ So, if $AB=BA,$ we have $$ B = \left( \begin{array}{cc} p & q \\ 0 & p \end{array} \right) $$ and $$ B = qA + (p-7q)I $$
Another way to say the same condition is that the characteristic and minimal polynomials are the same.
I collected several variations on this, I will go find it... Given a matrix, is there always another matrix which commutes with it?