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Let V be an n-dimensional inner product space over C and T ∈ L(V ). Suppose that $T^2 = T^∗T$. Prove that T is self-adjoint.

My idea is since $T^*T$ is positive, then all the eigenvalues of it should be nonnegative. Also, since $T^2=T^*T$, then this is also true for $T^2$. Then suppose $T$ has a eigenvalue $\lambda$, then $T^2$ should have eigenvalue $\lambda^2\geq0$. This forces all the eigenvalues for $T$ must be real, since it's impossible to have $(a+bi)^2\geq0$ with $b\neq 0$. Thus, now I think can prove some other things to prove $T$ is self-adjoint.

  1. If $T$ is normal, then $T$ is self-adjoint

  2. By Schur theorem, there exists an orthonormal basis such that $T$ is upper triangular. If I can prove all the entries above the diagonal of $M(T)$ are zero, then $T$ is self-adjoint

However, unfortunately, I can prove none of them. Also, I was able to solve a similar problem

If $TT^*=T^2$ then $T$ is self-adjoint

This question enable me to say that if $Tv=0$ then $T^*v=0$ by using norm $$ \|T^*v\|^2=\langle T^*v,T^*v\rangle=\langle TT^*v,v\rangle=\langle T^2v,v\rangle=0$$

And by $\dim$ null $T$ = $\dim$ null $T^*$, I can prove null $T$ = null $T^*$ which gives $T$ is normal, but this seems not useful in this problem, or probably I missed something

Anyway, any help is appreciated. Thank!

M_k
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    try using cauchy-schwarz – user8675309 Apr 06 '22 at 00:54
  • $\langle T^Tv,v\rangle =|Tv|^2$ ( since $T^T$ is positive, so no need for absolute value). Then $\langle T^2v,v\rangle =\langle Tv,T^v\rangle \leq |Tv||T^v|$ Since $T^T=T^2$, this gives $|Tv|^2 \leq |Tv| |T^v| $. Thus if $T^*v=0$, then this forces $Tv =0 $ Is this right? – M_k Apr 06 '22 at 01:11

1 Answers1

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If you were able to prove

$$\tag1TT^*=T^2\ \implies\ T=T^*$$

and you have $T^*T=T^2$, then taking adjoints you have $T^*T=(T^*)^2$. Letting $S=T^*$, this is $$ SS^*=S^2.$$ Now you know by $(1)$ that $S=S^*$, so $T=T^*$.

Using more explicitly that you are in a finite-dimensional setting, here is another argument. You have $$ (T-T^*)^2=T^2+T^{*2}-T^*T-TT^*=T^*T-TT^*. $$ Multiplying by $i$ and taking the trace, $$ \operatorname{Tr}\big[[i(T-T^*)]^2\big]=0 $$ So the operator $i(T-T^*)$ is selfadjoint and its square is zero; so it is zero. Then $T-T^*=0$.

Martin Argerami
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  • Thanks! The (1) is exactly what I was trying to do but I ended up confusing myself... – M_k Apr 06 '22 at 02:23
  • Also, could you please explain why multiplying by $i$? And $i(T-T^)^2=iT^T-iTT^$, and if I take the adjoint of it, the result will be $-iT^T+iTT^*$, so is this really self-adjoint? – M_k Apr 06 '22 at 02:30
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    No, its $i(T-T^)$ that is selfadjoint. Its square is $TT^-T^*T$. – Martin Argerami Apr 06 '22 at 03:12
  • So you mean that $i(T-T^)$ is self-adjoint, so it only has real eigenvalues $\lambda$, then the square of it is $[i(T-T^)]^2$ which has eigenvalues $\lambda^2$. Then since the trace of its square is zero, $\sum \lambda^2=0$, and hence this implies all $\lambda$ is zero, which gives $i(T-T^)$ has an orthonormal eigenbasis with all eigenvalues are zero. Thus $i(T-T^)=0\implies T=T^*$. This is the right interpretation? – M_k Apr 06 '22 at 04:14
  • Yep. Exactly that. – Martin Argerami Apr 06 '22 at 04:24