Why do we assume that a matrix in quadratic form is Symmetric?
Solution 1:
Any matrix A can be written as sum of $(A+A^{T})/2$ and $(A-A^{T})/2$. You can verify that the quadratic form of second term (i.e; $x^{T}(A-A^{T})x$) turns out to be zero (Try to evaluate the second term and check for yourself).
Hence, we can assume that we begin with a symmetric matrix. If not, it is anyway possible to convert to an equivalent quadratic form with a symmetric matrix.
Solution 2:
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 $${\bf x}^\top {\bf M} {\bf x} = {\bf x}^\top {\bf M}_s {\bf x}$$
Solution 3:
The main reason for getting the matrix of a real quadratic form symmetric by replacing the original matrix with its symmetric part ${A+A^T}\over 2$ is that any symmetric matrix is orthogonally diagonalizable and all eigenvalues are real. Then for symmetric $A$ you have some orthogonal matrix $U$ (that is, $U^T=U^{-1}$) and $U^T A U=D$ is real diagonal, and $x^T A x=(Ux)^T(UAU^T)(Ux)=(Ux)^TD(Ux)$ is a sum of squares with real coefficients where the length $\Vert x \Vert$ of the vector $x$ is the same as the length $\Vert Ux\Vert$ of the vector $Ux$.