Prove for any matrix $B$, we have $y^{\top}By = y^{\top}Ay$ with $A$ symmetric

Question

I need to handle the following question:

Prove for any matrix $B$, we have $y^TBy = y^TAy$ with $A$ symmetric.

Now, we can of course simply pick $B$ to be $A^T$, but that would be pretty trivial.

Does anyone have a tip or a hint how to start? I thought we might consider decomposing the symmetric matrix into two seperate matrices (symmetric and skew symmetric), but I don't know whether that helps

Answer 1

Note that for any square matrix $B$ we may define

$B_{sym} = \dfrac{B + B^T}{2}, \tag 1$

and

$B_{skew} = \dfrac{B - B^T}{2}; \tag 2$

we have

$B_{sym}^T = \dfrac{B^T + (B^T)^T}{2} = \dfrac{B + B^T}{2} = B_{sym}, \tag 3$

and

$B_{skew}^T = \dfrac{B^T - (B^T)^T}{2} = \dfrac{B^T - B}{2} = -B_{skew}, \tag 4$

and

$B_{sym} + B_{skew} = \dfrac{B + B^T}{2} + \dfrac{B - B^T}{2} = \dfrac{B + B^T + B - B^T}{2} = \dfrac{2B}{2} = B. \tag 5$

and

$B_{sym} - B_{skew} = \dfrac{B + B^T}{2} - \dfrac{B - B^T}{2} = \dfrac{B + B^T - B + B^T}{2} = \dfrac{2B^T}{2} = B^T; \tag 6$

therefore, since $y^TB_{skew}y$ is a scalar quantity

$y^TB_{skew}y = (y^TB_{skew}y)^T = y^TB_{skew}^T(y^T)^T = -y^TB_{skew}y, \tag 7$

which evidently implies

$y^TB_{skew}y = 0 \tag 8$

over any field $\Bbb F$ with

$\text{char}(\Bbb F) \ne 2; \tag 9$

in such cases we thus conclude that

$y^TBy = y^T(B_{sym} + B_{skew})y= y^TB_{sym}y + y^TB_{skew}y=y^TB_{sym}y; \tag{10}$

now by virtue of (3) we may take

$A = B_{sym}. \tag{11}$