I'm trying trying to prove diagonalizable criterion for operators and matrices. It's a little lengthy:

Let $V$ have basis $\{e_1,...,e_n\}$ and fix $T\in\scr{L}$$(V)$. I know for any matrix $A(T)$ of $T$, the $\lambda$ is an eigenvalue of $A(T)\iff\lambda$ is an eigenvalue of $T$. Also, $a_1e_1+...+a_ne_n=:v\in V$ is an eigenvector of $T\iff (a_1,...,a_n)\in\mathbb{F}^n$ is an eigenvector of $A(T)$. In particular, we have the following: Since dim$V=$dim$\mathbb{F}^n$, $V\cong\mathbb{F}^n$. Let $\phi$ be the isomorphism $a_1e_1+...a_ne_n\mapsto (a_1,...,a_n)$ between them. Isomorphisms carry linearly independent sets to linearly independent sets. Hence, $\{v_{\lambda_1},...,v_{\lambda_n}\}$ is a basis of eigenvectors of $T\iff \{\phi(v_{\lambda_1}),...,\phi(v_{\lambda_n})\}$ is a basis of eigenvectors of $\mathbb{F}^n$.


1.) It's clear that $T$ has a diagonal matrix $\iff$ $V$ has a basis of eigenvectors of $T$.

2.) If $T$ has $n$ distinct eigenvalues, then $\{v_{\lambda_1},...,v_{\lambda_n}\}$ is a basis (of eigenvectors) of $V$. Hence, $T$ has $n$ distinct eigenvalues $\implies T$ has a diagonal matrix


3.) $A(T)$ is diagonalizable $\iff$ $A(T)\sim D(T)$ (diagonal matrix) $\iff T$ has basis of eigenvectors of $V\iff$ $A(T)$ has a basis of eigenvectors of $\mathbb{F}^n$, those eigenvectors being precisely as described in the first paragraph.

4.) $A(T)$ has $n$ distinct eigenvalues $\implies$ $A(T)$ is diagonliazable. Proof: $A(T)$ has $n$ distinct eigenvalues $\iff T$ has $n$ distinct eigenvalues (by first paragraph) $\implies$ $T$ has basis of eigenvectors of $V$ (by 2) $\iff A(T)$ has basis of eigenvectors of $\mathbb{F}^n$ (by first paragraph) $\iff A(T)$ is diagonlizable (by 3).

Does this look ok?

  • 1
    $\begingroup$ Actually, the part with the "distinct eigenvalues" is either ambiguously worded or wrong. What you want is for the minimal polynomial to have distinct roots (all with multiplicity 1). The characteristic polynomial can contain the same root multiple times and so the diagonal matrix corresponding to $A$ can have repeated diagonal elements. Other than that it looks ok, but of course you didn't prove the "hard" parts (namely why 2 holds). $\endgroup$ – Fryie Dec 2 '15 at 10:24
  • $\begingroup$ For the distinct eigenvalues part, do you mean #4? I only wanted to say that part is true in one direction, I know the converse isn't necessarily true. Is that what you meant? $\endgroup$ – user153582 Dec 2 '15 at 10:39
  • $\begingroup$ Ah yes, sorry, I missed that. In that case it's right. $\endgroup$ – Fryie Dec 2 '15 at 10:46
  • $\begingroup$ Cool. Thanks for checking :) $\endgroup$ – user153582 Dec 2 '15 at 10:49

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