# Proof Regarding Diagonalizability, Eigenspace and Multiplicity

A linear operator $T$ on a finite-dimensional vector space $V$ over $\mathbb{R}$. Let $a_1, \dots, a_k$ be distinct eigenvalues. Let $m_i$ be the multiplicity of $a_i$ as a root of the characteristic polynomial of $T$. Then $T$ is diagonalizable if and only if

1. $m_1 + \dots + m_k = n = \dim(V)$, and

2. for each $i$, $\dim(E_{a_i}) = m_i$, where $E_{a_i}$ denotes the eigenspace of the corresponding eigenvalue.

==============================

I have proved $\impliedby$ direction (by assuming two conditions hold and prove the diagonalizability of $T$)

However, For $\implies$ direction, which is assuming $T$ is diagnoalizable and prove conditions (1) and (2). I don't where should I start

Compute the characteristic polynomial of a diagonal matrix. Then argue that if $T$ is diagonalizable, then it any diagonal matrix representing it have the same characteristic polynomial, same eigenvalues, same everything.
• Here is my thought: If $T$ is diagonalizable, then there is a basis $\beta$={$v_1,...v_n$} consisting of eigenvectors for $T$. Since the dimension of a eigenspace is less than or equal to its corresponding multiplicity. Thus, dim($E_{\lambda_i}$)$\leq$$m_i. If we let l_i be the number of eigenvectors for \lambda_i in \beta (also the number of the basis vectors for eigenspace E_{\lambda_i}, thus l_i=dim(E_{\lambda_i}) ) We have l_i$$\leq$$m_i$ for all $i$. Commented Aug 24, 2014 at 23:26
• Also because $\Sigma{l_i}$= n (the dimension of $V$) and $\Sigma{m_i}\leq n$, this implies $\Sigma{l_i}=\Sigma{m_i}=n$ and $l_i=m_i$ for each $i$. Because for each $i$ we have $l_i$=dim($E_{\lambda_i}$), therefore, $l_i$=dim($E_{\lambda_i}$)=$m_i$. However, by computing charateristic polynomial, do you mean that I need use determinants? Commented Aug 24, 2014 at 23:26
• and I also not sure if $\Sigma{m_i}\leq n$ is correct Commented Aug 24, 2014 at 23:30