# Inner product space defined on unitary space

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.

• 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 $a_{11} = e_1^T A e_t = \langle e_1, e_1 \rangle > 0$. – Joppy May 2 '18 at 6:11
• Yes you are right. But, I want to verify whether the norm defined is correct or not. @Joppy – Atul Anurag Sharma May 2 '18 at 6:14
• 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. – Joppy May 2 '18 at 6:43
• By correct or not I mean whether it is well defined or not. @Joppy – Atul Anurag Sharma May 2 '18 at 8:25
• 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. – Joppy May 2 '18 at 8:35

$\langle x,x \rangle \geq 0$ for all $x$ and this implies that $A$ is a non-negative definite matrix. All non-negative definite matrices have the stated properties.
• Yes. On any Complex Hilbert space all operators with $\langle Tx, x \rangle \geq 0$ for all $x$ are Hermitian ($\equiv$ self-adjoint). – Kabo Murphy May 3 '18 at 10:42