2
$\begingroup$

Let $A \in \mathbb C^{n \times n}$ be such that $<x,y> = y^{\theta} Ax$ is an inner product on $\mathbb C^n$. Prove that A is a hermitian matrix with positive diagonal entries.

The first problem I am facing is to verify the properties of the inner product space defined above. Also, I need some help regarding how to prove A is a hermitian matrix with positive entries on its diagonal.

$\endgroup$
  • $\begingroup$ I don't think you're meant to verify the properties of an inner product space, the question is saying that $A$ is chosen specifically such that $\langle x, y \rangle := y^T A x$ is an inner product. For example, this will imply that the upper left entry is real and positive, since $a_{11} = e_1^T A e_t = \langle e_1, e_1 \rangle > 0$. $\endgroup$ – Joppy May 2 '18 at 6:11
  • $\begingroup$ Yes you are right. But, I want to verify whether the norm defined is correct or not. @Joppy $\endgroup$ – Atul Anurag Sharma May 2 '18 at 6:14
  • $\begingroup$ What do you mean by correct or not? The matrix $A$ is chosen in such a way that $y^T A x$ is an inner product. $\endgroup$ – Joppy May 2 '18 at 6:43
  • $\begingroup$ By correct or not I mean whether it is well defined or not. @Joppy $\endgroup$ – Atul Anurag Sharma May 2 '18 at 8:25
  • $\begingroup$ As it is defined there, the bracket $\langle -, - \rangle$ is clearly $\mathbb{R}$-bilinear, $\mathbb{C}$-linear in the first argument, and $\mathbb{C}$-conjugate-linear in the second argument. So $\langle -, - \rangle$ is well-defined as a sesquilinear form. The question then asserts that it is an inner product, and you need to find the conditions on $A$ which make this true. $\endgroup$ – Joppy May 2 '18 at 8:35
0
$\begingroup$

$\langle x,x \rangle \geq 0$ for all $x$ and this implies that $A$ is a non-negative definite matrix. All non-negative definite matrices have the stated properties.

$\endgroup$
  • $\begingroup$ Are you saying that all positive semidefinite matrices are Hermitian? $\endgroup$ – Joppy May 2 '18 at 12:26
  • $\begingroup$ Yes. On any Complex Hilbert space all operators with $\langle Tx, x \rangle \geq 0$ for all $x$ are Hermitian ($\equiv$ self-adjoint). $\endgroup$ – Kabo Murphy May 3 '18 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.