4
$\begingroup$

Here is a question from Axler's Linear Algebra Done Right (Chapter 6.B problem 11).

Suppose $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$ are inner products on $V$ such that $\langle v,w\rangle_1 = 0$ if and only if $\langle v,w \rangle_2 = 0$. Prove that there is a positive number $c$ such that $\langle v,w\rangle_1 = c\langle v,w\rangle_2$ for every $v,w \in V$.

It doesn't say that $V$ is finite dimensional, but I tried showing the result assuming this. If $\{e_1,\dots,e_n\}$ is an orthonormal basis with respect to $\langle \cdot,\cdot \rangle_1$, then $\langle e_i,e_j \rangle_1 = 0$ for all $i \neq j$. This implies $\langle e_i,e_j \rangle_2 = 0$ for all $i \neq j$. So we can use the equivalence they give to show that a set of vectors is orthogonal with respect to one inner product if and only if they are orthogonal with respect to the other. But I'm not sure how to get the result from here.

Another idea I had was to represent the linear functional $\langle \cdot,w \rangle_1$ using Riesz Representation. There exists some $u \in V$ such that $\langle \cdot,w\rangle_1 = \langle \cdot, u \rangle_2$. That is, for any $v\in V$, we have \begin{align*} \langle v,w \rangle_1 = \langle v,u \rangle_2 \end{align*} If we want $\langle v,u \rangle_2 = c\langle v,w\rangle_1$ for some positive $c$, then we would have $c = \frac{\langle v,u \rangle_2}{\langle v,w\rangle_1}$, but I'm not sure how to show this is a positive constant.

$\endgroup$
  • 1
    $\begingroup$ Hint: $\langle x, y - \frac{\langle x, y \rangle_1}{\langle x, x \rangle_1} x \rangle_1 = 0$. If you change 1 to 2 on the outermost inner product, then expand, you can get a relation from which you can bootstrap to the full result. $\endgroup$ – Daniel Schepler Jun 21 '17 at 0:37
  • $\begingroup$ Per your instruction, I expand the expression and get $ \frac{\langle x,x \rangle_1}{\langle x,x \rangle_2} \langle x,y \rangle_2 = \langle x,y \rangle_1$. So we have $c = \frac{\langle x,x \rangle_1}{\langle x,x \rangle_2}$. But this $c$ depends on $x$, and we want $c$ to hold for all $x,y \in V$. $\endgroup$ – Daniel Xiang Jun 21 '17 at 0:43
  • 1
    $\begingroup$ So, as another way of putting it, $\frac{\langle x, y \rangle_2}{\langle x, y \rangle_1} = \frac{\langle x, x \rangle_2}{\langle x, x \rangle_1}$. Now, what happens if you reverse the roles of $x$ and $y$? $\endgroup$ – Daniel Schepler Jun 21 '17 at 0:45
  • $\begingroup$ Brilliant. Thank you. $\endgroup$ – Daniel Xiang Jun 21 '17 at 0:52

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.