Does the statement $x^\top Ax \cdot x^\top Bx = |x|^2 x^\top ABx$ hold for symmetric $A,B$?

Question

I have the following statement: $x^\top Ax \cdot x^\top Bx = |x|^2 x^\top ABx$, both $A$ and $B$ are symmetric matrices. It is trivial to prove it if $AX = XA$, where $X = xx^\top$. But is it correct in general case?

Answer 1

No. It is not true.

Hint to find a counter example:

Let $x$ be some standard unit vector.

Answer 2

Take $A=B$ diagonal. It becomes the "identity" $$\left(\sum_i x_i^2\right)\left(\sum_i x_i^2a_{ii}^2\right)=\left(\sum_i x_i^2 a_{ii}\right)^2$$ If such an identity were true, you'd know it from high school, if not earlier. However, no one ever teaches anything like this, and for good reason.

Answer 3

I think it does not! Check for example the case:

$$A = B = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$$

The left hand side gives $x^\top Ax \cdot x^\top Bx = 4 x_1^2 x_2^2$ while the right hand side gives $|x|^2 x^\top ABx = |x|^4 = (x_1 + x_2)^4$.