The Wikipedia article for Hermitian matrices gives an alternative characterization for Hermitian matrices link. The statement I want to prove is: $$ A^* = A \iff \langle w,Av \rangle_H = \langle Aw,v \rangle_H $$ This was proven for the standard inner product by a similar question. I want to prove it for the general case with an arbitrary hermitian inner product on complex vector spaces. I tried it using $$ \langle w,Av \rangle_H = w^*HAv $$ with $H$ being a hermitian matrix and $ * $ marks the hermitian conjugate of a vector/matrix. I don't really know how to show it without using that H and A commute.

I am thankful for any help!


Note that I am using a positive definite hermitian form as the definition for the inner product. A hermitian form is defined as a sesquilinear form with $\langle w,v \rangle = \overline{\langle v,w \rangle}$.

Then there exists a hermitian matrix H being the transformation matrix of the hermitian form with $$ \langle w,v \rangle_H = w^*Hv $$

  • $\begingroup$ Just to be clear: your definition of the inner product is $\langle x, y\rangle_H=x^* H y$? $\endgroup$ Jun 18 at 21:04
  • $\begingroup$ @DavidRaveh yes exactly, I added it to the question. $\endgroup$ Jun 18 at 21:13
  • $\begingroup$ Your definition of the inner product isn't valid, since it doesn't generally satisfy $\langle x,x \rangle_H\geq 0$ $\endgroup$ Jun 18 at 21:22
  • $\begingroup$ @DavidRaveh you're right. I clarified it a bit. please see the edit in the question. $\endgroup$ Jun 18 at 21:46

1 Answer 1


The difference between this proof and the one you linked is mostly symbolic. We define the adjoint of $A$ denoted by $A^\dagger$ by $\langle A^\dagger i|j\rangle=\langle i|Aj\rangle$.

For any inner product, if $A=A^\dagger$ then $$A_{ij}=\langle i|Aj\rangle=\langle Ai|j\rangle=\langle j|Ai\rangle^*=A_{ji}^*$$ so $A$ is Hermitian. The converse follows by going in the opposite direction as well.

  • $\begingroup$ I think I don't quite get it then. Isn't $\langle Ax, y\rangle_H = (Ax)^*y$ (copied from the other question) Implying that $ H = I $ with $ I $ being the identity matrix. Or is it that the basis can be chosen to fulfil $ H=I $? $\endgroup$ Jun 18 at 22:17
  • $\begingroup$ @exploitingDev I didn't realize that they didn't use general notation, so I edited to redo it. If you have any questions about how I defined the adjoint, we can go to a chatroom $\endgroup$ Jun 18 at 22:47
  • $\begingroup$ Thanks for your help. It makes sense now. $\endgroup$ Jun 19 at 7:59

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