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Let $A$, $B$ be complex matrices that commute. Show that if they are $2$x$2$, then there is a polynomial $p$ such that either $p(A)=B$ or $p(B)=A$. Show the result is false for $3$x$3$.

I'm not really sure where to start with this problem. The only way I could think of $2$x$2$ being relevant, was that you know exactly what the coefficients of the characteristic polynomials are (in terms of trace and determinant). But this doesn't seem to help. Can someone provide a hint how to get start? Thank you.

user124910
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  • have you studied Jordan form? – Will Jagy Aug 15 '16 at 02:53
  • @WillJagy Yes, I have – user124910 Aug 15 '16 at 02:59
  • OK, second question: Did you learn anything special about two matrices that commute, related to their Jordan decomposition? –  Aug 15 '16 at 03:03
  • @mathguy Off the top of my head, I don't believe so. Which result are you referring to? – user124910 Aug 15 '16 at 03:06
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    None (sorry I was trying to make a certain point indirectly). Did you learn anything special about two matrices that commute, related to putting them in triangular form? (Hint: It can be done simultaneously). Also note that if $X$ is an invertible matrix, then solving the problem for $A$ and $B$ is equivalent to solving it for $XAX^{-1}$ and $XBX^{-1}$. So... you may assume the matrices are triangular. Does that help? –  Aug 15 '16 at 03:12
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    @mathguy http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 – Will Jagy Aug 15 '16 at 03:15
  • @mathguy I see. Yes, that does help a lot. Thank you – user124910 Aug 15 '16 at 03:17

1 Answers1

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I would divide this into cases.

Case 1: One matrix is a multiple of the identity. We can simply use a constant polynomial, then.

Case 2: One matrix (WLOG call it $A$) is diagonalizable with two distinct eigenvalues $\lambda_1,\lambda_2$.

We can show that since $B$ commutes with $A$, is simultaneously diagonalizable with eigenvalues $\mu_1,\mu_2$. It suffices to find a $p$ such that $p(\lambda_i) = \mu_i$ for $i=1,2$.

Case 3: $A$ is not diagonalizable.

Without loss of generality (after an appropriate change of basis), suppose that $A$ is in Jordan form, so that $$ A = \pmatrix{\lambda & 1\\0 & \lambda} $$ Note that $B$ commutes with $A - \lambda I$. Conclude that $B$ has the form $$ B = \pmatrix{\mu & t\\0 & \mu} $$ Write $B = \mu I + t(A - \lambda I)$.

For the $3 \times 3$ case, consider the diagonal matrices $$ A \pmatrix{1\\&1\\&&0}, \quad B = \pmatrix{0\\&1\\&&1} $$

Ben Grossmann
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