Let $d(\lambda_i)$ represent the geometric multiplicity, and let $m(\lambda_i)$ represent the algebraic multiplicity.
Theorem: Let $T$ be a linear operator on an $n$-dimensional vectorspace. Then $T$ is diagonalizable if and only if for each eigenvalue $\lambda_i$, we have $d(\lambda_i) = m(\lambda_i)$.
I proved $\Leftarrow$ already. I'm reading a book where the converse is proven, but I'm having trouble understanding it. It goes as follows:
Suppose $T$ is diagonalizable. Then $V$ has a basis of eigenvectors of $T$. Let $\beta$ be such a basis. Write the vectors of $\beta$ in such a way that they are ordered according to the corresponding eigenvalue: the first $m_1$ vectors correspond with the first eigenvalue $\lambda_1$, the next $m_2$ vectors correspond with the second eigenvalue $\lambda_2$, etc. It suffices to construct the diagonal matrix of $T$ with respect to this basis to see that $d(\lambda_i) = m(\lambda_i)$ voor alle eigenvalues $\lambda_i$.
I'm not sure if this really is a proof, and I don't understand what he is trying to say with his last assertion, i.e. that constructing the diagonal matrix proves what we want. Can someone clarify this please, or prove this part in an alternative manner?
Best Answer
Suppose that $T$ is diagonalizable, then there exist a basis $B$ of $V$ such that $[T]_B$ is a diagonal matrix with its diagonal elements as eigenvalues of $T$, so that characteristic polynomial splits over $F$. Since, $$V=E_{\lambda_1}\oplus E_{\lambda_2}\oplus...\oplus E_{\lambda_k}$$ so if $B_i$ is basis of $E_{\lambda_i}$, then $$B=\bigcup_{i=1}^k B_i $$ is a basis of $V$ and $[T]_B$=diag($[T_1]_{B_1},...,[T_k]_{B_k})$ where, $T_i$ is a linear operator on $E_{\lambda_i}$ induced by $T$.
Let dim$(E_{\lambda_i})=n_i$, since $T_i(x)=T(x)=\lambda_i(x)$ for all $x\in E_{\lambda_i}$ so that $$[T_i]_{B_i}=\lambda_i I_{n_i}$$, Hence the characteristic polynomial is,
$$(x-\lambda_1)^{n_1}(x-\lambda_2)^{n_2},...,(x-\lambda_k)^{n_k}$$
Which shows that $AM=GM$ for each $\lambda_i$.