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Is the following true with a non standard inner product?

$\langle M\cdot x1, x2\rangle = \langle x1, M^\mathsf{T}\cdot x2\rangle$

$x1,x2 \in \mathbb{C}^n$

$M$ is an $n \times n$ square complex matrix.

$M^\mathsf{T}$ is its conjugate transpose.

It is true for the standard inner product but it is not true with a different inner product.

For example, for the following inner product, the equation is not true.

$\langle v1, v2\rangle := 2.0 \cdot v1[0]\cdot v2[0] + v1[1] \cdot v2[1] + v1[2] \cdot v2[2] +...$

(Only the multiply of the first row is doubled, and the rest are same)

Is it true only with the standard inner product?

Jihyun
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1 Answers1

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I'll denote the hermitian transpose / conjugate transpose of $M$ by $M^H$. Note that you might write any inner product in the form $\langle u, v\rangle = v^H A u$ with some complex matrix $A$ with entries $a_{ij} = \langle e_i, e_j \rangle$. Let's assume that $\langle Bu, v \rangle = \langle u, B^H v \rangle$ for all matrices $B$, then we have $$ \langle B e_i, e_j \rangle = e_j A B e_i = \sum_{k=1}^n a_{jk}b_{ki} \\ \overset{!}{=} \langle e_i, B^H e_j \rangle = e_j^H BA e_i = \sum_{k=1}^n b_{jk} a_{ki} $$ where $e_j$ are the elements of the canonical basis. So in words: $A$ has to commute with any matrix $B$. As can be shown the only matrices statisfying this property are matrices of the form $\lambda I_n$ where $I_n$ is the identity matrix and $\lambda \in \Bbb C$ (see for example A linear operator commuting with all such operators is a scalar multiple of the identity.).

So no, this is true for infinitely many dot products - but they're not really that interesting.

In general however you will always find some matrix $B^*$ such that $\langle Bu, v\rangle = \langle u, B^* v \rangle$. These are extremely interesting and are called adjoint matrices (or adjoint functions / maps / operators / endomorphisms etc.) and basically generalize this behaviour of the hermitian transpose with the standard dot-product.

SV-97
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  • Thank you very much. So ⟨,⟩=⟨,∗⟩ is true only when ∗=. When the inner product is the standard one, is and thus commutes with any matrix. just needs to be a hermitian (∗=). – Jihyun Oct 13 '21 at 09:21
  • I tried look for a when = $\begin{bmatrix}2&0&0\0&1&0\0&0&1\end{bmatrix}$, which is the in my question. In that case, is a hermitian with some 0s. \begin{bmatrix}a&0&0\0&b&d\0&d*&c\end{bmatrix} – Jihyun Oct 13 '21 at 09:34
  • There's a minor detail you gotta pay attention to: my answer is about the case where this shifting property holds for any matrix $B$, whereas in your example it holds for this specific $B$ but if you for example consider $$ B = \begin{pmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 2 & 0 & 0 \end{pmatrix} $$ then $AB \neq BA$. You will of course always find some matrix (in fact you'll find infinitely many, since they form a vector space) such that the multiplication commutes (e.g. the trivial case where $A=B$). – SV-97 Oct 13 '21 at 13:17