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Let V be a finite dim complex inner product space. Let T be a linear operator on V. Prove that T is self-adjoint if and only if $\langle T\alpha,\alpha\rangle$ is real $\forall \alpha \in V$. The first direction is trivial, if we assume T is self-adjoint then $\langle T\alpha,\alpha\rangle = \langle \alpha,T\alpha\rangle=\overline{\langle T\alpha,\alpha\rangle}$ So it has to be real. But I have no clue on how to start the other direction? Where would I start?

ultrainstinct
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  • You may think $T$ as a matrix. Now consider $T = (T + T^)/ 2 + (T - T^)/(2i) i$. – user251257 Apr 04 '16 at 14:43
  • So here would you just replace T in $\langle T\alpha,\alpha\rangle$ with that. Then use linearity in the first argument to split it up, and do something with that? – ultrainstinct Apr 04 '16 at 14:48
  • yes. Just figure out which terms are real – user251257 Apr 04 '16 at 14:50
  • I still can't figure it out. I wrote it down and I am trying to manipulate the inner product using its properties, but nothing is working to show that it is self-adjoint. Why does figuring out which terms are real help with anything? – ultrainstinct Apr 04 '16 at 15:17

2 Answers2

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Hint: Let $\{e_1,\dots,e_n\}$ be an orthonormal basis vectors of $V$. It suffices to show that the matrix of $T$ with respect to this basis is Hermitian.

Note that $\langle T e_j, e_j \rangle$ is real. What does this tell you about the diagonal entries?

Note that $\langle T(e_j + e_k),(e_j + e_k)\rangle$ and $\langle T(e_j + ie_k),(e_j + ie_k)\rangle$ are real. What does that tell you about the rest of the entries?

Ben Grossmann
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Let $A = (T + T^*) / 2$ and $B = (T - T^*) / (2i)$. Notice that $A$ and $B$ are both self adjoint and we have $T = A + Bi$.

Now, for every $\alpha$ $$ \langle T \alpha, \alpha\rangle = \langle A \alpha, \alpha\rangle + \langle B \alpha, \alpha\rangle i $$ is real. Thus, $\langle B \alpha, \alpha\rangle$ is zero for every $\alpha$, but that implies $T = T^*$.

user251257
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