REVISTED$^1$: Circumstantial Proof: $P\implies Q \overset{?}{\implies} Q\implies P$ 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._ !enter image description here !enter image description here ## EDIT$^1$: If $A$ is similar to a diagonlizable matrix, then $A$ is diagonalizable, right?

No. It would only prove the _converse_.

$$\underbrace{P \implies Q}_{\text {implication}} \quad\
ot\equiv \underbrace{Q \implies P}_{\text{converse of implication}}$$

If you need to prove $P \implies Q$, you **can** prove its equivalent:

$$\underbrace{\lnot Q \implies \lnot P}_{\text{contrapositive of implication}}$$