To prove that if a matrix $A\in M_{n\times n} ( F )$ has $n$ distinct eigenvalues, then $A$ is diagonalizable is enough to show that the opposite holds? That is, if $A$ is diagonalizable, then $A$ has distinct eigenvalues?
Please don't answer the actual question. I want to do it myself. I just want to know if it suffice to show the opposite in this particular case.
If $A$ is similar to a diagonlizable matrix, then $A$ is diagonalizable, right?
On
To help on this point of logic, let me try something I suggest to my students. You take the sentence
If it's raining, then the ground is wet.
as a reasonable one. And the contrapositive
If the ground is dry, then it is not raining.
seems equivalent and totally reasonable, as well. (So these are logically equivalent statements.)
However, the converse
If the ground is wet, then it is raining.
is certainly not believable. It may have rained hours ago. Or someone spilled some water.
Concrete examples may help you deal with the logical structure in the mathematical theorems.
No. It would only prove the converse.
$$\underbrace{P \implies Q}_{\text {implication}} \quad\not\equiv \underbrace{Q \implies P}_{\text{converse of implication}}$$
If you need to prove $P \implies Q$, you can prove its equivalent:
$$\underbrace{\lnot Q \implies \lnot P}_{\text{contrapositive of implication}}$$