1
$\begingroup$

I'm having trouble understanding the adjoint of a linear operator in a Hilbert space. I have the following problem but I'm unsure of my answers:

Let A and B be bounded linear operators mapping a complex Hilbert space H into itself. Prove that:

a) $(\alpha A + \beta B)^*=\bar\alpha A^* + \bar\beta B^*$

b) $(AB)^* = B^*A^*$

c) $(A^*)^* = A$

d) $I^* = I$, where $I$ is the identity operator.

So I believe I just use the inner product, using the Riesz representation theorem to find a specific vector $y$ such that $\langle Ax, y\rangle = \langle x, A^*y\rangle$. So my solution for the first one (the rest are similar) is:

\begin{align*}\langle(\alpha A+\beta B)x, y\rangle &= \alpha\langle Ax, y\rangle + \beta \langle Bx, y\rangle\\ &=\alpha\langle x, A^*y\rangle + \beta\langle x, B^*y\rangle\\ &=\langle x, \bar\alpha A^*y\rangle + \langle x, \bar\beta B^*y\rangle\\ &=\langle x, (\bar\alpha A^* + \bar\beta B^*)y\rangle, \end{align*} therefore $\bar\alpha A^* + \bar\beta B^* = (\alpha A + \beta B)^*$.

Is this the right idea? We use the inner product for the adjoint of an operator on Hilbert spaces into themselves because the Riesz representation theorem ensures we can always find a special $y$ to have the equality $\langle Ax, y\rangle = \langle x, A^*y \rangle$? We would not use the inner product for the adjoint if we were going between two different Hilbert spaces, correct?

$\endgroup$
1
$\begingroup$

That's the main idea. The argument could be made slightly more complete by adding a bit more:

For any $x,y\in H$, we have \begin{align*} \langle x,(\alpha A+\beta B)^*y\rangle&=\langle(\alpha A+\beta B)x,y\rangle\\&=\cdots \text{(exactly what you did)}\cdots\\ &=\langle x,(\overline\alpha A^*+\overline\beta B^*)y\rangle. \end{align*} Since $x$ and $y$ were arbitrary, we have $(\alpha A+\beta B)^*=\overline\alpha A^*+\overline\beta B^*$.

This uses the following lemma: If $H$ is a complex Hilbert space and $T\in\mathcal B(H)$, then $T=0$ iff $\langle Tx,y\rangle=0$ for all $x,y\in H$.

Yes, the Riesz representation theorem is used to guarantee the existence of the adjoint. For given $y\in H$, the map $x\mapsto\langle Tx,y\rangle$ is a bounded linear functional, hence there exists a unique $T^*y\in H$ such that $\langle Tx,y\rangle=\langle x,T^*y\rangle$ for all $x\in H$. Then showing the map $y\mapsto T^*y$ is linear and bounded is a basic exercise.

Furthermore, essentially the same proof works for $T:H_1\to H_2$. You still use the Riesz representation theorem to obtain a map $T^*:H_2\to H_1$. You just have to be careful and pay attention to what space the vectors are in.

$\endgroup$
2
  • $\begingroup$ Thank you for your help, I see I was mistaken when it comes to adjoints between two different Hilbert spaces. I will mess around with those to become more comfortable with the whole idea, but I feel comfortable in my solutions to this problem now-thank you again for all your help! $\endgroup$ – benty Apr 20 '18 at 22:12
  • 1
    $\begingroup$ You're welcome. Glad to help! $\endgroup$ – Aweygan Apr 20 '18 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.