Is the matrix $A$ symmetric in the quadratic form?

Question

Given $x \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ ($A$ is not necessarily symmetric), the quadratic form is written as $x^TAx$, a scaler. We have, $$x^TAx=(x^TAx)^T=x^TA^Tx$$ that is $x^T(A-A^T)x=0$

Why couldn't conclude $A=A^T$ from $x^T(A-A^T)x=0$, where $x \ne \boldsymbol{0}$? I know it's a false statement and there are counter examples, but it seems to me, mathematically, $A$ should be symmetric. Could someone help explain why I couldn't make such an inference?

Answer 1

If your question is:

Question: Given a matrix $A$ such that $$x^TAx=x^TA^Tx \, \forall x$$ does it follow that $A=A^T$?

The answer is no, a counterexample is $$A=\begin{bmatrix} 0 & 1\\-1 &0 \end{bmatrix}$$

In general, any quadratic form corresponds to infinitely many matrices. Among those, exactly one is symmetric.

Note Lets pick $A=\begin{bmatrix} a & b\\c &d \end{bmatrix}$ to be a $2 \times 2$ matrix. Then $$x^TAx=ax_1^2+dx_2^2+(b+c)x_1x_2$$

This means that the quadratic form only tells you what $a,d$ and $b+c$ are. You don't know what $b$ and $c$ are.

Answer 2

In fact, any quadratic form can be reduced to a symmetric matrix. Suppose $A$ is a non-symmetric matrix. Then $(A-A^T)/2$ is skew-symmetric since $$ \left(\frac{A-A^T}{2}\right)^T=\frac{A^T-A}{2}=-\frac{A-A^T}{2} $$ And for a skew-symmetric matrix it is always true that $$ x^T\left(\frac{A-A^T}{2}\right)x=0 $$ Since $$ x^T\left(\frac{A-A^T}{2}\right)x=\left(x^T\left(\frac{A-A^T}{2}\right)x\right)^T=x^T\left(\frac{A-A^T}{2}\right)^Tx=-x^T\left(\frac{A-A^T}{2}\right)x $$ Thus we have $$ x^TAx=x^T\left(\frac{A+A^T}{2}+\frac{A-A^T}{2}\right)x=x^T\left(\frac{A+A^T}{2}\right)x+x^T\left(\frac{A-A^T}{2}\right)x=x^T\left(\frac{A+A^T}{2}\right)x $$ Clearly, $(A+A^T)/2$ is symmetric. So a quadratic form matrix can be always reduced to a symmetric matrix.