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](https://i.stack.imgur.com/j2JPv.jpg)

![enter image description here](https://i.stack.imgur.com/bw4fz.jpg)

EDIT$^1$:
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If $A$ is similar to a diagonlizable matrix, then $A$ is diagonalizable, right?