I wish to show the following theorem:

Let $T:H\to H$ be a bounded linear operator on a complex Hilbert space $H$. Then if $\left\langle Tx,x\right\rangle \in\mathbb{R}$ for all $x\in H$, then $T$ is self-adjoint.

My proof is as follows:

If $\left\langle Tx,x\right\rangle \in\mathbb{R}$ for each $x\in H$ then for each $x \in \mathbb{R}$: $$ \left\langle Tx,x\right\rangle =\overline{\left\langle Tx,x\right\rangle }=\overline{\overline{\left\langle x,Tx\right\rangle }}=\left\langle x,Tx\right\rangle $$ so that $T$ is self-adjoint by definition. (For the first equality, I have used the fact that $x \in \mathbb{R} \Leftrightarrow \bar{x} = x$, for the second, I have used the general property in a Hilbert space that $<x,y> = \overline{<y,x>}$ and for the third that for all $x \in \mathbb{C}: \overline{\bar{x}} = x$.)

I'm satisfied with this proof, but the one I have in my textbook is more complicated, showing that $0 = <(T - T^{*})x, x>$ and a Lemma that says that if $Q : X \to X$ is a bounded linear operator on a complex inner product space $X$ and $ <Qx, x> = 0 \forall x \in X$ then $Q = 0$. Am I missing something subtle?

  • 2
    $\begingroup$ No. For a bounded linear operator, self-adjoint is: $(Tx,y)=(x,Ty)$ for every $x,y$. Note also that the result is only true in complex Hilbert spaces. False for real ones. $\endgroup$
    – Julien
    May 2, 2013 at 13:12
  • 2
    $\begingroup$ I believe this is what has been said already, but just to reiterate: You have not successfully shown that $T$ is self-adjoint. To show that $T$ is self-adjoint, you must instead show that $\langle Tx, y \rangle = \langle x, Ty \rangle$ for arbitrary $x$ and $y$. It does not suffice to show the equality for $x = y$. $\endgroup$
    – user73640
    May 2, 2013 at 13:19

1 Answer 1


Your argument does not prove the result at this point. For a bounded linear operator, $T$ is self-adjoint ($T^*=T$) iff $(Tx,y)=(x,Ty)$ for every $x,y\in H$. So you need to do more.

Using the textbook lemma is a good idea. It yields a straightforward argument. The proof of that lemma is not very difficult if you are familiar with polarization. It boils down to observing that $$ 2(Qx,y)=(Q(x+y),x+y)-i(Q(x+iy),x+iy)=0\qquad\forall x,y\in H. $$ Then for $y=Qx$, you get $(Qx,Qx)=\|Qx\|^2=0$, whence $Qx=0$ for every $x$, i.e. $Q=0$. So you just have to observe that $Q=T-T^*$ satisfies the condition $$(Qx,x)=(Tx,x)-(T^*x,x)=(Tx,x)-(x,Tx)=(Tx,x)-\overline{(x,Tx)}=(Tx,x)-(Tx,x)=0.$$ The lemma proves that $Q=T-T^*=0$, i.e. $T=T^*$ as desired.

You could also use polarization-like formulae to prove directly that $(Tx,y)=(x,Ty)$. But I think using this lemma is better.

Note that the lemma is false in the real case. Think of $\pi/2$ rotations. For instance, on $\mathbb{R}^2$: $$ Q=\pmatrix{0&-1\\1&0}\quad\Rightarrow\quad (Qx,x)=-x_2x_1+x_1x_2=0\quad\forall x\in\mathbb{R}^2. $$


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