To represent a bilinear operator on finite dimensional space by a matrix

Question

This exercise somehow connects functional analysis to linear algebra. However, I am still not be able to fully understand the insights behind it.

Recall the standard inner product on $\mathbb{C}^n$ for $x=(x_1,x_2,...,x_n)^T$ and $y=(y_1,y_2,...,y_n)^T$ defined by \begin{align} \langle x|y \rangle := \sum_{j=1}^n \overline{x_j} \; y_j \end{align}

1. Use a basis on $\mathbb{C}^n$ to show that any map $T:\mathbb{C}^n \times \mathbb{C}^n \to \mathbb{C}$ that is linear in the second variable and conjugate linear in the first variable is of the form \begin{align} T(x,y) := \sum_{j,k=1}^n A_{j,k} \; \overline{x_j} \; y_k \end{align} for a matrix $A = (A_{j,k})_{1\le j,k \le n} \in \mathbb{M}_n(\mathbb{C})$.

2. Show that the matrix $A$ above is self-adjoint if and only if $T(x,y) = \overline{T(y,x)}$ for all $x,y \in \mathbb{C}$.

3. Show that $A$ is positive semi-definite if and only if $T(x,x) \ge 0$ for all $x \in \mathbb{C}^n$ and it is positive definite if and only if, in addition, $T(x,x) = 0$ implese $x = 0$.

I am trying to do the exercise above, as follows:

We define a basis $\{v_j:j=1,2,...,n\}$ for $\mathbb{C}^n$ such that $T(v_j,v_k) := (A_{j,k})_{1\le j,k \le n}$, then $x = \sum_{j=1}^n x_j v_j$ and $y = \sum_{k=1}^n y_k v_k$ there holds \begin{align} T(x,y) &= T(\sum_{j=1}^n x_j v_j,\sum_{k=1}^n y_k v_k) \\ &= \sum_{j=1}^n \overline{x_j} \; T(v_j,\sum_{k=1}^n y_k v_k) \\ &= \sum_{j,k=1}^n \overline{x_j}\;y_k \; T(v_j,v_k) \\ &= \sum_{j,k=1}^n \overline{x_j}\;y_k \; A_{j,k} \end{align} which is in the required form. Is this a solution? I am not sure if (and why) we can always define such a basis. Is this basis always orthonormal?

$A$ is self-adjoint if and only if $\langle Ax|y \rangle = \langle x|Ay \rangle$, that is, \begin{align} \sum_{j,k=1}^n \overline{A_{k,j}\;x_j} \; y_k = \sum_{j,k=1}^n \overline{x_j} \; A_{j,k}\;y_k \end{align} which implies $A_{j,k} = \overline{A_{k,j}}$, and thus $T(v_j,v_k) = \overline{T(v_k,v_j)}$ by the definition of $T$. Due to the linearity of $T$ applied to $x = \sum_{j=1}^n x_j v_j$ and $y = \sum_{k=1}^n y_k v_k$, we have $T(x,y) = \overline{T(y,x)}$ for all $x,y \in \mathbb{C}$.

Any comments on my solution? As for the 3rd question anyone may help me?

Answer 1

(1). You don't need to find a basis that suffices $T(v_i,v_j) = A_{ij}$. What you need is to take any orthonormal basis (for example, $v_i=(0,...,\underbrace{1}_{\text{$i$-th place}},...0)^T$). And define matrix $A_{ij}=T(v_i,v_j)$. Then do what you did.

(3). Depends on your definition of positive (semi-)definite matrix. In some books, it's that $A_{ij}v_iv_j > 0$ ($\ge 0$) for all non-zero $v_i$, which is what you basically have, since you have shown (1).