In an inner product space $V$ with inner product $\langle\cdot,\cdot\rangle$ with a second inner product $g$ on $V$ , why does there exists an self-adjoint operator such that

$$\langle x,y\rangle= g(T(x),y)$$

for all $x,y\in V$?

  • 1
    $\begingroup$ Try to show this result in the case where $g$ is a standard scalar product $g(x,y)=\sum_j x_j \bar y_j$ and then it will be easy to generalise. $\endgroup$ Aug 17, 2016 at 16:13
  • $\begingroup$ Is your space finite- or infinite-dimensional? Which inner product is $T$ supposed to be self-adjoint with respect to? $\endgroup$ Aug 18, 2016 at 15:40

1 Answer 1


To make the problem a bit more symmetric, suppose that $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on the same finite-dimensional space $V$. Choose an orthonormal basis $\{ e_n \}_{n=1}^{N}$ of $V$ with respect to the second inner product. Then $\langle e_n,e_{n'}\rangle_2 = \delta_{n,n'}$, and \begin{align} \langle x,y\rangle_1 & = \left\langle x, \sum_{n=1}^{N}\langle y,e_n\rangle_2 e_n\right\rangle_1 \\ & = \sum_{n=1}^{N}\langle e_n,y\rangle_2 \langle x,e_n\rangle_1 \\ & = \sum_{n=1}^{N}\langle x,e_n\rangle_1 \langle e_n,y\rangle_2 \\ & = \left\langle \sum_{n=1}^{N}\langle x,e_n\rangle_1 e_n,y \right\rangle_2 \end{align} So, the linear operator $T : V\rightarrow V$ defined by $$ Tx = \sum_{n=1}^{N}\langle x,e_n\rangle_1 e_n. $$ satisfies $$ \langle x,y\rangle_1 = \langle Tx,y\rangle_2 $$

  • $\begingroup$ Is this normally how one compares inner products? Why is the conjecture a self-adjoint operator? $\endgroup$ Aug 20, 2016 at 13:23
  • $\begingroup$ If the inner products are complex, for example, then $\langle Tx,y\rangle_2 = \langle x,y\rangle_1 = \overline{\langle y,x\rangle_1}=\overline{\langle Ty,x\rangle_2}= \langle x,Ty\rangle_2$. Similarly, you can show $T^{-1}$ is selfadjoint on $V$ with respect to inner product $1$. $\endgroup$ Aug 20, 2016 at 13:38
  • $\begingroup$ I meant why is the conjecture that "to compare 2 inner products, we conjecture that the difference is by a self-adjoint map $T$" $\endgroup$ Aug 20, 2016 at 13:39
  • $\begingroup$ @jacobsmith : I forgot to ping you in my comment. $\endgroup$ Aug 20, 2016 at 13:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .