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It is for homework. But I am not asking the answer, since it seems to be already given in many posts like this one... I am pretty sure that I will get an A just copying and pasting it.

My question is a little bit about logic, I think.

Many texts argue that two matrices A and B commute IF AND ONLY IF they share a not null set of eigenvectors.

In a quick research about logic I found that $P \iff Q$ has the following truth table:

enter image description here

So, calling events:

$E_1$ : matrices A and B share a not null set of eigenvectors

$E_2$ : matrices A and B commute

I understand that this proof requires at least two steps:

  1. $E_1 \implies E_2$
  2. $\neg E_1 \implies \neg E_2$

The first part seems to be fulfilled by:

For a $\mathbf{A}, \mathbf{B} \in \mathbb{C}^n \times \mathbb{C}^n$, say $\{x_i\} \in \mathbb{C}^n$ is a set of vectors such that:

$\mathbf{A} {x_i} = a_i {x_i}$, where $a_i$ is a eigenvalue of $\mathbf{A}$

$\mathbf{B} {x_i} = b_i {x_i}$, where $b_i$ is a eigenvalue of $\mathbf{B}$

Then we can see that:

$\mathbf{B}(\mathbf{A} x_i) = \mathbf{B}(a_i x_i) = a_i(\mathbf{B} x_i) = a_i b_i x_i$

$\mathbf{A}(\mathbf{B} x_i) = \mathbf{A}(b_i x_i) = b_i(\mathbf{A} x_i) = a_i b_i x_i$

I see it in in much more precise shapes and in many places.

How to show the second part? Is it necessary right?

Given $\mathbf{A}, \mathbf{B} \in \mathbb{C}^n \times \mathbb{C}^n$

If, for some $y \in \mathbb{C}^n$:

$\nexists \lambda \in \mathbb{C} | \mathbf{A} y = \lambda y$

Then:

$\mathbf{B} (\mathbf{A} y) \neq \mathbf{A} (\mathbf{B} y)$

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    $\begingroup$ Traditionally if you have a proposition $P\iff Q$ you prove $P\implies Q$ and $Q\implies P$ rather than $\neg P \implies \neg Q$. However, that is just the contrapositive of the second proposition that you need to prove, so it would be a valid proof if you could prove it that way. $\endgroup$ – Doug M Mar 21 '18 at 21:13
  • $\begingroup$ @DougM, thanks for your comment. It makes sense. I was not seeing this way. I try this out. Thank you. $\endgroup$ – Thiago Melo Mar 21 '18 at 21:25

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