A diagonalizable matrix's proof I came a cross the following question today at class

If a matrix has $n$ eigenvalues and it is known all of them are different from each other then $A$ is a diagonalizable matrix 

to best of my knowledge, if the eigenvalues are different from each other (and let's say you got $n$ of those), you can make $n$ bases vectors using those eigenvalues, but I'd love to see a proper proof :)
cheers
 A: Here's a constructive approach: you know that
$$
A v
= \lambda v
$$ for corresponding eigenvectors $v$ and eigenvalues $\lambda$. Now write
$$
S := (v_1, \ldots, v_n)
\qquad \text{and} \qquad
D := \text{diag}(\lambda_1, \ldots, \lambda_n),
$$
i.e. the columns of $S$ are the eigenvectors and $D$ is a diagonal matrix containing the eigenvalues on its diagonal.
As the eigenvectors are linearly independent (verify this if you haven't already, it's a good exercise!) you can invert $S$.
Try to prove that $$A = S D S^{-1}.$$
Spoiler:

 Rearrange the above equation to $A S = S D$, which is equivalent to $A v_i = \lambda_i v_i$ for all $i \in \{1, \ldots, d\}$.

A: If the set of eigenvalues is $\{\lambda_1,\lambda_2,\ldots,\lambda_n\}$ then, for each $j\in\{1,2,\ldots,n\}$, let $v_j$ be an eigenvector corresponding to the eigenvalue $\lambda_j$. Then, since eigenvectors corresponding to distinct eigenvalues are linearly independent, the set $\{v_1,v_2,\ldots,v_n\}$ is a basis of your space. Then $A$ is similar to$$\begin{bmatrix}\lambda_1&0&\ldots&0\\0&\lambda_2&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&\lambda_n\end{bmatrix}.$$
