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 $A.$

  • $\begingroup$ What exactly is your definition of the adjoint $A^*$? Are you just taking $A^*$ to mean the conjugate transpose matrix of the matrix $A$? $\endgroup$ Feb 13 '20 at 4:03
  • $\begingroup$ Yes, that's right $\endgroup$ Feb 13 '20 at 4:16
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 13 '20 at 5:49
  • $\begingroup$ Yes, I completely agree! Well, can this be proved then? $\endgroup$ Feb 13 '20 at 5:50
  • $\begingroup$ Standard notation is not $<Au,Av>=\cdots,$ but $\langle Au,Av\rangle=\cdots. \qquad$ $\endgroup$ 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*}

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

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