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The general form of an inner product in $\mathbb{C}^n$ is $\langle x,y\rangle=y^{*}Bx$ where B is a Hermitian positive definite matrix. Then for any square matrix $A$ we have $\langle Av,w\rangle=w^{*}BAv$ and $\langle v,A^{*}w\rangle=(A^{*}w)^{*}Bv=w^{*}ABv$.

Since for any matrix $A$, it is a fact that $\langle Av,w\rangle=\langle v,A^{*}w\rangle$ then we have $w^{*}BAv=\langle Av,w\rangle=\langle v,A^{*}w\rangle=w^{*}ABv$. But that implies $BA=AB$, which is wrong. What am I doing wrong?

epsilon
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    If by $A^\ast$ you denote the conjugate transpose, you only have $\langle Av, w\rangle = \langle v, A^\ast w\rangle$ for all matrices $A$ and vectors $v,w$ if $\langle\cdot,\cdot\rangle$ is (a multiple of) the standard inner product. For other inner products, the adjoint of $A$ is not the conjugate transpose. – Daniel Fischer Oct 11 '13 at 19:04
  • Then, are you saying that in http://www.math.wisc.edu/~jensen/Algebra/inner.pdf the lemma in the second page is wrong? – epsilon Oct 11 '13 at 19:12
  • Supposedly, they consider only the standard inner product. But they could also be plain wrong, I haven't looked. – Daniel Fischer Oct 11 '13 at 19:18
  • They assume the inner product is any inner product. This is made clear by the proposition following that lemma where they relate any inner product to the standard inner product. – epsilon Oct 11 '13 at 19:24
  • Okay, then I will have to take a look. Give e a few minutes. – Daniel Fischer Oct 11 '13 at 19:28

3 Answers3

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That is because $A^*$ is defined with respect to the standard inner product (that is, when $B=I$). So in your case you do not have $(Ax, y) = (x, A^*y)$.

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It's an error in your source.

With an inner product $\langle\cdot,\cdot\rangle_B$ given by a (hermitian positive definite) matrix $B$, i.e.

$$\langle v, w\rangle_B = w^\ast\cdot B \cdot v,$$

for any matrix $A$, there is an adjoint matrix $\tilde{A}^B$ such that for all $v, w$ we have

$$\langle Av, w\rangle_B = \langle v, \tilde{A}^B w\rangle_B.$$

Written out,

$$w^\ast B A v = (\tilde{A}^Bw)^\ast B v = w^\ast \left(\tilde{A}^B\right)^\ast B v.$$

For that to hold for all $v,w$, the matrices in the middle must be the same,

$$BA = \left(\tilde{A}^B\right)^\ast B \iff BAB^{-1} = \left(\tilde{A}^B\right)^\ast \iff \tilde{A}^B = (BAB^{-1})^\ast = B^{-1} A^\ast B.$$

The $B$-adjoint matrix is in general not the conjugate transpose.

Daniel Fischer
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  • Thank you Daniel, I am surprised that the same mistake appears in several books. For example in http://www.amazon.com/Elementary-Functional-Analysis-Graduate-Mathematics/dp/0387855289/ref=sr_1_1?ie=UTF8&qid=1381541003&sr=8-1&keywords=barbara+maccluer on page 35. Can you recommend any good reliable books on this topic? Also when is the matrix B a diagonal matrix and when is not? Thank you – epsilon Oct 12 '13 at 02:47
  • @epsilon Can't see page 35 of that. But I have a theory. If you consider an endomorphism $f$ and its adjoint, then the matrix of $f^\ast$ with respect to an orthonormal basis $\mathcal{F}$ is the conjugate transpose of the matrix of $f$ with repsect to $\mathcal{F}$. But then the matrix $B$ of the inner product with respect to the basis $\mathcal{F}$ is the identity matrix. I suspect that is stated in the linked book, but it's not stated so in the PDF. – Daniel Fischer Oct 13 '13 at 01:07
  • Daniel, I haven't seen that stated in that book. That's why I was asking for a recommendation for another book. As you do, I believe that with respect to an orthonormal basis the matrix B has to be the identity so the adjoint of the transformation is the conjugate transpose of the matrix giving the transformation. – epsilon Oct 13 '13 at 23:53
  • Sorry, I don't know any books on linear algebra (where that rightly belongs). The books on Functional Analysis that I'm more or less familiar with all assume the reader to be familiar with that, hence don't go into it. I think Lang has a book on linear algebra, that would then most likely be a good one, but of course only if you appreciate Lang's style. – Daniel Fischer Oct 14 '13 at 00:02
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This isn't true; other people have explained the reasoning, you have to take the adjoint relative to $B$. Here's an example:

$$ B = \left(\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right), A = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right) $$

$B$ is Hermitian positive definite, but

$$ \left<Ax,y\right> = y^*BAx = y^*\left(\begin{array}{cc} 0 & 2 \\ 0 & 1 \end{array}\right)x\\ \left<x,A^*y\right> = y^*ABx = y^*\left(\begin{array}{cc} 1 & 2 \\ 0 & 0 \end{array}\right)x\\ $$ and it's not hard to find $x,y$ so that these aren't equal ($x = y = (1,0)^T$ e.g.).

BaronVT
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  • Thank you BaronVT. I thought the matrix B was always diagonal. When is B diagonal and when it's not? Thank you – epsilon Oct 12 '13 at 02:55
  • You can induce an inner product with any Hermitian, positive-definite matrix. Obviously, diagonal positive-definite is sufficient (i.e. a diagonal matrix with positive entries), but not necessary.

    But even in this case, diagonal matrices $B$ will not commute with arbitrary $A$ (unless $B = cI$), and so the adjoint still has to be taken relative to $B$.

     – BaronVT Oct 14 '13 at 15:02