2
$\begingroup$

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $i$. $\endgroup$
    – ElleryL
    Commented Aug 24, 2014 at 23:26
  • $\begingroup$ 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? $\endgroup$
    – ElleryL
    Commented Aug 24, 2014 at 23:26
  • $\begingroup$ and I also not sure if $\Sigma{m_i}\leq n$ is correct $\endgroup$
    – ElleryL
    Commented Aug 24, 2014 at 23:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .