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

• Note that what you have is basically all the essential ingredients. That's the hard part. All that's lacking is a bit of formalism to make it completely correct. – Arthur Jan 19 at 12:54
• You might add that the matrix is $n\times n$. – amd Jan 19 at 18:09
• – Martin Sleziak Jan 19 at 21:34

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 $$A = S D S^{-1}.$$
Rearrange the above equation to $$A S = S D$$, which is equivalent to $$A v_i = \lambda_i v_i$$ for all $$i \in \{1, \ldots, d\}$$.
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$$\begin{bmatrix}\lambda_1&0&\ldots&0\\0&\lambda_2&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&\lambda_n\end{bmatrix}.$$