Proof of Theorem DED

Let $\scalarlist{\lambda}{n}$ denote the $n$ distinct eigenvalues of $A$. Then by Theorem NEM we have $n=\sum_{i=1}^n\algmult{A}{\lambda_i}$, which implies that $\algmult{A}{\lambda_i}=1$, $1\leq i\leq n$. From Theorem ME it follows that $\geomult{A}{\lambda_i}=1$, $1\leq i\leq n$. So $\geomult{A}{\lambda_i}=\algmult{A}{\lambda_i}$, $1\leq i\leq n$ and Theorem DMFE says $A$ is diagonalizable.