# Why can an arbitrary operator be linearized with a suitable basis?

(Wu-Ki Tung, Group theory in physics, in proving the first Schur lemma) Let $A$ an arbitrary operator in space $V$. It can be made to be hermitian*, and a basis of the space can be chosen to consist of the eigenvectors $A$, i.e.

$$Au_{\alpha,i}=\lambda_iu_{\alpha,i}$$

Why does this always happen?

*Without loss of generality, we can take $A$ to be hermitian (if it's not, then it can be decomposed into two hermitian operators $A_+=(A-A^\dagger)/2, A_-=(A-A^\dagger)/2i$, consider each of them separately then combining them in $A=A_++iA_-$

• I would look at the first 15 or so pages of Serre, Linear Representations of Finite Groups. – D_S Nov 25 '17 at 3:57
• This is not true without more hypotheses on $A$; consider, for example, $A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$ acting on $\mathbb{R}^2$. – Qiaochu Yuan Nov 25 '17 at 4:10
• @QiaochuYuan so A can't be any arbitrary operator? – Ooker Nov 25 '17 at 4:12
• No. But you don't need a statement like this to prove Schur's lemma anyway, at least some versions of it. – Qiaochu Yuan Nov 25 '17 at 4:15
• How do you make it Hermitian? – user7530 Nov 27 '17 at 2:43