Showing the bilinear and commutative map $f:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ should be in the quadratic form.

Question

I'm trying to show, any map $f:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ that holds, for $x,y,x_1,x_2 \in \mathbb{R}^n, c \in \mathbb{R}$,

• $ f(x,y)=f(y,x)$
• $ f(x_1+x_2,y) = f(x_1, y) + f(x_2, y)$
• $ f(cx, y) = cf(x,y)$

should be expressed in the quadratic form $f(\mathbf{x}, \mathbf{y}) = \mathbf{x}^T A \mathbf{y}$, where $A$ is symmetric.

My attempt

It can be shown that the quadratic form map $f(\mathbf{x}, \mathbf{y}) = \mathbf{x}^T A \mathbf{y}$ satisfies,

• $f(\mathbf{x}, \mathbf{y}) = \mathbf{x}^T A \mathbf{y} = (\mathbf{x}^T A \mathbf{y})^T = f(\mathbf{y}, \mathbf{x})$

• $f(\mathbf{x_1} + \mathbf{x_2}, \mathbf{y}) = f(\mathbf{x_1}, \mathbf{y}) + f(\mathbf{x_2}, \mathbf{y})$

• $f(c \mathbf{x}, \mathbf{y}) = cf( \mathbf{x}, \mathbf{y})$

But I'm stuck in proving the opposite direction.

Answer 1

Hint: Let $v_i$ be an orthonormal basis. Define $f(v_i,v_j)=a_{ij}$ Now check that the matrix $A=(a_{ij})$ does the trick by testing it again on the orthonormal basis $v_i$.