TITLE: Why do we assume that a matrix in quadratic form is Symmetric?
QUESTION [16 upvotes]: I am looking to the review document for linear algebra (Zico Kolter (updated by Chuong Do), Linear Algebra Review and Reference), and the part of the quadratic form  (pg17) mentions about an assumption of being symmetric for a matrix in quadratic form. It also includes some declarative equality for that proposed argument. 

What is the practical reason to assume that the matrix describing a quadratic form (over $\mathbb{R}$) is symmetric? I also do not get the idea proposed by the argument? Can someone light me about?

REPLY [9 votes]: We can show that the Matrix $\bf{W}$ of the quadratic form $x^T {\bf{W}} x$ can be assumed to be symmetric in the following way:
In general a matrix can be decomposed into a symmetric and an anti-symmetric part as follows:
$$M = \frac{1}{2} ( {\bf{M}} + {\bf{M}}^\top) + \frac{1}{2} ({\bf{M}} - {\bf{M}}^\top) = M_s + M_a$$
Further we note that the symmetric part is invariant to the transpose,
$$ {\bf{M}}_s^\top = {\bf{M}}_s $$
while the antisymmetric part changes its sign under transposition:
$$ {\bf{M}}_a^\top = - {\bf{M}}_a$$
Let us examine the quadratic form of a purely antisymmetric matrix:
$$q = {\bf x}^\top {\bf M}_a {\bf x} = ({{\bf x}^\top {\bf M}_a^\top {\bf x}})^\top  = -({\bf x}^\top {\bf M}_a {\bf x})^\top = -q$$
Which implies that $q=0$ has to hold true.
Now for a general Matrix $\bf W$ we can see that:
$${\bf x}^\top {\bf M} {\bf x} = {\bf x}^\top ({\bf M}_s + {\bf M}_a){\bf x} = {\bf x}^\top {\bf M}_s {\bf x} + {\bf x}^\top {\bf M}_a {\bf x}$$
As we know that $${\bf x}^\top {\bf M}_a {\bf x} = 0$$ we finally find
$${\bf x}^\top {\bf M} {\bf x} = {\bf x}^\top {\bf M}_s {\bf x}$$