Inner Product on $\mathbb C^d$ Induces Hermitian Matrix

Question

Let $\left\langle\cdot,\cdot\right\rangle$ be an inner product on $\mathbb C^d$. I want to show that there exists a matrix $A$ such that $\left\langle x,y\right\rangle=Ax\cdot y$ and $Ax\cdot x>0$ for any nonzero $x$. Could someone please provide ideas?

Answer 1

Let $e_i$ denote the $i$th standard basis vector. If such a matrix $A$ existed, then we must have $$\langle e_j, e_i \rangle = A e_j \cdot e_i = A_{i,j}.$$ You can then check that defining $A$ in this way will give you what you want.

Answer 2

Hint: Ensure the desired property for the standard basis elements first.

Answer 3

Let (e_i) be a basis and let u_ij := , U=(u_ij), then = x^T Uy and x^T U x. Then A is the conjugate transpose of U, i.e A = U^*.

Answer 4

Fix $y \in \mathbb{C}^d$. The map $x \mapsto \langle x, y\rangle$ is a linear functional on $\mathbb{C}^d$ so by the Riesz representation theorem there exists $\tilde{y} \in \mathbb{C}^d$ such that $\langle x, y\rangle = x \cdot \tilde{y}, \forall x\in \mathbb{C}^d$.

Define a map $B : \mathbb{C}^d \to \mathbb{C}^d$ with $By = \tilde{y}$. We claim that $B$ is linear.

$$\langle x, \alpha y + \beta z\rangle = \overline{\alpha} \langle x, y\rangle + \overline{\beta}\langle x,z\rangle = \overline{\alpha} x \cdot \tilde{y} + \overline{\beta}x \cdot \tilde{z} = x\cdot (\alpha \tilde{y} + \beta \tilde{z}), \forall x\in \mathbb{C}^d$$

Therefore $\widetilde{\alpha y + \beta z} = \alpha \tilde{y} + \beta \tilde{z}$.

We conclude $$\langle x, y\rangle = x \cdot By = (B^*x) \cdot y$$

so the desired linear map is $A = B^*$.

In fact, this proof directly generalizes to a result on Hilbert spaces:

Theorem.

Let $H$ be a Hilbert space with the inner product $\langle\cdot, \cdot\rangle$ and let $[\cdot, \cdot]$ be another inner product on $H$. Then there exists a bounded linear map $A : H \to H$ such that $[x,y] = \langle Ax, y\rangle, \forall x,y \in H$.

We just need to check that $B$ is bounded. Assume $y_n \xrightarrow{n\to\infty} y$ and $\tilde{y_n} \xrightarrow{n\to\infty} z$.

We have

$$\langle x, z\rangle \xleftarrow{n\to\infty}\langle x, y_n\rangle = [x,y_n] \xrightarrow{n\to\infty} [x,y] = \langle x, \tilde{y}\rangle, \quad\forall x \in H$$

so $z - \tilde{y} \perp H$ which implies $z = \tilde{y}$. By the Closed Graph Theorem we conclude that $B$ is bounded.

Now we can take the adjoint and set $A = B^*$.