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Actual Question is:

Let $\mathcal{F}$ be a commuting family of $3\times 3$ complex matrices. How many linearly independent matrices can $\mathcal{F}$ contain? what about the $n\times n$ case? (Hoffman Kunzze, Linear algebra, 6.5.2)

I have no idea ow to go directly for $n\times n$.. SO, I thought i would try for $2\times 2$ and $3\times 3$ and then generalize.

For $2\times 2$ I know that basis of $\mathcal{M_2}=\{2\times 2 ~complex ~ matrices\}$ is $ \left( \begin{array}{cccc} 1 & 0 \\ 0 & 0 \end{array} \right),\left( \begin{array}{cccc} 0 & 1 \\ 0 & 0 \end{array} \right),\left( \begin{array}{cccc} 0 & 0 \\ 1 & 0 \end{array} \right) ,\left( \begin{array}{cccc} 0 & 0 \\ 0 & 1 \end{array} \right) $

So, what i was thinking is if i can check which matrices commutes in this connection, this would be the linearly independent commuting sets of $2\times 2$ matrices ( this is my guess not very sure)

we have $ \left( \begin{array}{cccc} 1 & 0 \\ 0 & 0 \end{array} \right)\left( \begin{array}{cccc} 0 & 1 \\ 0 & 0 \end{array} \right)= \left( \begin{array}{cccc} 0 & 1 \\ 0 & 0 \end{array} \right)$ but, $ \left( \begin{array}{cccc} 1 & 0 \\ 0 & 0 \end{array} \right)\left( \begin{array}{cccc} 0 & 1 \\ 0 & 0 \end{array} \right)=\left( \begin{array}{cccc} 0 & 0 \\ 0 & 0 \end{array} \right)$

i dont want to write all other combinations, but, i would write only commuting matrices

$\left( \begin{array}{cccc} 1 & 0 \\ 0 & 0 \end{array} \right)\left( \begin{array}{cccc} 0 & 0 \\ 0 & 1 \end{array} \right)=\left( \begin{array}{cccc} 0 & 0 \\ 0 & 0 \end{array} \right)$ and $\left( \begin{array}{cccc} 0 & 0 \\ 0 & 1 \end{array} \right)\left( \begin{array}{cccc} 1 & 0 \\ 0 & 0 \end{array} \right)=\left( \begin{array}{cccc} 0 & 0 \\ 0 & 0 \end{array} \right)$

So, I think these two elements must be linearly independent in the set of commuting $2\times 2$ matrices.

I am expecting to do the same for general $n\times n$ but this would be cumbersome..

So, I would be thankful if some one can help me out to solve this in detail... atleast for $2\times 2$ ...

Thank You.

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  • $\begingroup$ Here's a hint. Note that if $p$ is a polynomial, then matrix $A$ will commute with $p(A)$. $\endgroup$ – hardmath Sep 30 '13 at 14:25
  • $\begingroup$ If one is diagonalizable, then they are all simultaneously diagonalizable, so that should give you atmost $n$ linearly independent matrices. Now try similar cases when they are not necessarily diagonalizable. $\endgroup$ – Prahlad Vaidyanathan Sep 30 '13 at 14:26
  • $\begingroup$ @PrahladVaidyanathan : I am sorry, I did not understand... $\endgroup$ – user87543 Sep 30 '13 at 14:27
  • $\begingroup$ @hardmath : Yes, I see that.. But, not so sure how does that help.. $\endgroup$ – user87543 Sep 30 '13 at 14:28
  • $\begingroup$ For a given matrix $A$, the dimension of $\{p(A)| p \text{ polynomial } \}$ is limited by the minimal polynomial of $A$. This should get you pretty close to maximizing the number of linearly independent commuting matrices. $\endgroup$ – hardmath Sep 30 '13 at 15:20
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From what I can gather, this seems like a hard problem to pin down an exact bound for, however I suspect the book just wants you to notice that all these matrices will be simultaneously triangulable. Linear independence does not change when conjugating by an invertible matrix (this is what I alluded to in my comment above), so the dimension of this space is $\leq$ the dimension of all upper triangular matrices; which is ...

[Also see large sets of commuting linearly independent matrices, and the link therein to a theorem of Schur.

Very interesting indeed]

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  • $\begingroup$ In fact, I was recently told by a professor that a slight tweaking of this problem was an open question. $\endgroup$ – Alex Youcis Sep 30 '13 at 14:49
  • $\begingroup$ What is the tweak? $\endgroup$ – Prahlad Vaidyanathan Sep 30 '13 at 14:52
  • $\begingroup$ I think it was to find a bound on the dimension of the algebra generated by $A_1,\ldots,A_m$ where $A_1,\ldots,A_m$ are commuting $n\times n$ matrices over a field. I could be misremembering though. If you are really interested, I could try and find out for you. $\endgroup$ – Alex Youcis Sep 30 '13 at 14:55
  • $\begingroup$ Nah, don't bother. It was just idle curiosity. $\endgroup$ – Prahlad Vaidyanathan Sep 30 '13 at 14:56
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I followed the link posted by Prahlad Vaidyanathan that gives a general (simplified) proof of Shcur's theorem which gives the exact answer for any $n$.

Based on that proof I don't know what Hoffman and Kunze could have possibly had in mind for the general case because it's non-trivial. The case $n=3$ may have a simpler proof that Hoffman and Kunze did have in mind but I have not been able to come up with anything simpler than extracting that proof as a special case of the simplified proof of Schur's theorem.

I typed up the proof in the case $n=3$ (which incidentally required proving it also for $n=2$). The LaTeX involved was sufficiently complicated that I've been unable to translate it into MathJax. So I'm displaying it as an image.

I would not have bothered doing this except that Hoffman and Kunze is such a timeless classic that it seems there should be better solutions available online. This at least fills one hole.

You can click on the following to make it bigger. Click once to bring it up in its own window and you might need to click again to enlarge it. But it is high res you should be able to read it.

Hoffman and Kunze Exercise 6.5.2

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