Friedberg linear algebra theorem 5.9 It said $T$ (on vector space V) is diagonalizable if and only if the multiplicity of $\lambda_i$ is equal to dim($E_{\lambda_i}$) for all $i$.
It sets $m_i$ denote the multiplicity of $\lambda_i$, $d_i$=dim($E_{\lambda_i}$), n=dim(V).
When prove the only if part (assuming $m_i = d_i$). It said $\sum_{i=1}^{k} m_i = n$, I don't understand why the sum of multiplicities equal to the dimension of the vector space. Any help? Thanks
 A: The characteristic polynomial $p$ of $T$ has degree $n$. On the other side we have
$p(\lambda)= (\lambda- \lambda_1)^{m_1}(\lambda- \lambda_2)^{m_2}...(\lambda- \lambda_k)^{m_k}$, where
$\lambda_1,...,\lambda_k$ are the distinct eigenvalues of $T$. Can you take it from here ?
A: It's not so easy to understand what exactly you meant to ask, but I'll give it a try:


*

*Using your notation, observe that it is always true that $\;\sum\limits_{i=1}^k m_i=n\;$, since we're summing up the algebraic multiplicities of all the roots of a polynomial of degree $\;n=\dim V\;$ , namely: the characteristic polynomial.

*Now, the claim you write at the beginning of your post says a square matrix (or a linear operator) $\;T\;$ on $\;V\;$ is diagonalizable iff $\;m_i=d_i\;$ , for all $\;i=1,2,...,k\;$ , but by point (1) above this is equivalent to require that $\;\sum\limits_{i=1}^k m_i=\sum\limits_{i=1}^k d_i=n\;$ ...
Resuming: You can either check that $\;m_i=d_i\;$ for each and every root of the characteristic polynomial  (i.e., for each and every eigenvalue of $\;T\;$ ), or you can check that $\;\sum\limits_{i=1}^k d_i=n\;$ . It is the same...though many times it is easier to check the first condition, i.e.: $\;d_i=m_i\;$ for each $\;i=1,2,...,k\;$ .
