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.
Thanks!!
Best Answer
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 ...