How can I prove the following:

An $n\times n$ matrix $A$ that has $n$ distinct eigenvalues is similar to a diagonal matrix.

I saw another question that was similar but it was about $n$ distinct eigenvectors, not eigenvalues, and I don't know if that will change the proof.



Let $\lambda_1,\ldots,\lambda_n$ be distinct eigenvalues. By definition, there exist corresponding eigenvectors $v_1,\ldots, v_n$, such that $v_i\ne 0$ and $Av_i=\lambda_iv_i$. As the eigenvalues are distinct, the $v_i$ are linearly independent and hence form a basis. Expressing $A$ in that base obviously produces a diagonal matrix ...


If all the eigen values are distinct then all the eigen vectors are distinct. Indeed let $\lambda_1\neq\lambda_2$ two eigen values, let $x_1$ an eigenvector of $\lambda_1$, let's show that $x_1$ can't be an eigenvector of $\lambda_2$.

If $x_1$ is also an eigenvector of $\lambda_2$, then : $Ax_1=\lambda_2 x_1\Rightarrow \lambda_1 x_1=\lambda_2 x_1\Rightarrow x_1(\lambda_1-\lambda_2)=0\Rightarrow \lambda_1-\lambda_2=0$. Because $x_1 \neq0$ because it is an eigenvector. So $\lambda_1=\lambda_2$, it is a contradiction.

So $A$ has $n$ different eigenvectors and you can conclude withe the result you mentionned.


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