# Proof of a Property of Inner Product Spaces

Is it true that $$\langle Au,Av\rangle = \langle u,A^{*}Av \rangle$$

for some inner product space $$V,$$ vectors $$u,v$$ and a square matrix $$A$$?

For the standard inner product, this is trivially true if $$A$$ is Hermitian, but what about other cases?

What if the inner product is not standard?

What if $$A$$ is not Hermitian?

When does the above hold true?

$$A*$$ refers to the conjugate transpose of $$A.$$

• What exactly is your definition of the adjoint $A^*$? Are you just taking $A^*$ to mean the conjugate transpose matrix of the matrix $A$? Feb 13 '20 at 4:03
• Yes, that's right Feb 13 '20 at 4:16
• Since your space is finite dimensional there is essentially only one inner product on it: any other inner product gives a space that is isometrically isomorphic to the one with the usual Euclidean inner product. Feb 13 '20 at 5:49
• Yes, I completely agree! Well, can this be proved then? Feb 13 '20 at 5:50
• Standard notation is not $<Au,Av>=\cdots,$ but $\langle Au,Av\rangle=\cdots. \qquad$ Feb 13 '20 at 6:48

As mentioned in the comments, on a finite dimensional space $$V$$ you have essentially the euclidean inner product. This means that any inner product on $$V$$ can be written as

$$\langle u, v \rangle_M = u^\ast M v$$

for a hermitian positive definite matrix $$M$$ (see e.g. here). But $$(V, \langle \cdot, \cdot \rangle_M)$$ is isometrically isomorphic to $$(V, \langle \cdot, \cdot \rangle_E)$$, where $$\langle \cdot, \cdot \rangle_E$$ denotes the euclidean inner product. To see this note that $$M$$ can be written as a square of a hermitian matrix $$M = N \cdot N.$$ Then $$(V, \langle \cdot, \cdot \rangle_M) \to (V, \langle \cdot, \cdot \rangle_E); \quad x \mapsto Nx$$ gives an isometric isomporphism, as $$\langle u, v\rangle_M = u^\ast M v = u^\ast NN v = \langle Nu, Nv \rangle_E.$$

Now we show the desired property for the euclidean inner product, which is

$$\langle u, v \rangle = u^\ast v$$ for arbitrary $$u, v \in V.$$

Thus \begin{align*} \langle Au, Av \rangle = (Au)^\ast Av = u^\ast A^\ast Av = \langle u, A^\ast Av \rangle. \end{align*}

• Could you show that there's only one type of inner product? Feb 14 '20 at 5:56
• Yes, edited a bit.
– blat
Feb 14 '20 at 7:47