0

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.

1 Answers1

0

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?

Community
  • 1
Will Jagy
  • 139,541
  • I see. Well, truth be told, I got this question from an old test. I double checked it, and I didn't miss anything. That is how it appeared. I guess someone made a mistake somewhere. Thanks for your help. No wonder I couldn't get it to work. –  Jul 14 '15 at 22:34
  • @fxy The test question must have been the converse "Assume that $A$ has distinct eigenvalues. Prove that $AB=BA$ implies $B=f(A)$" (which is the same as to say that $B$ is in the span above). It is the standard test question in the matrix theory courses. – A.Γ. Jul 14 '15 at 23:08
  • @A.G. Actually, it wasn't the converse (not that it matters too much). In reality, the question was posed as a "The following are equivalent" type. I was just having trouble with the direction above (if you read my post, I said I thought I proven the converse; this is why). Someone legitimately messed it up. It's okay though, these things happen. I've certainly encountered mistakes in exams before, as I'm sure others have as well. Thanks for your comment, though. –  Jul 14 '15 at 23:10
  • @fxy I see. The correct statement would be "$AB=BA$ implies $B=f(A)$ iff the Jordan blocks of $A$ has distinct eigenvalues". The later is equivalent to the minimal polynomial is the characteristic polynomial (as Will Jagy mentioned above). – A.Γ. Jul 14 '15 at 23:54