Proof One Can Convert Non-Symmetric Square Matrix Into a Symmetric Square Matrix I've read in "Introduction to Optimization" by Chong Et. Al that a non-symmetric square matrix can always be written as a symmetric square matrix. (pp.26)
How does he know this? Is there a proof for this?
Here's the passage.
 A: Only from the viewpoint of the quadratic form being defined. If $F$ is a skewsymmetric matrix, then we always have $x^TFx = 0$ when $x$ is a column vector. There are some simple tricks involved. First, a one by one matrix is regarded as a number. Next, the transpose of a one by one is itself. So, 
$$ (x^T F x)^T = x^T F x, $$
$$  x^T F^T x = x^T F x,$$
$$  x^T (-F) x = x^T F x,$$
$$  - x^T F^T x = x^T F x, $$
$$ 0 = 2 x^T F x,$$
$$ x^T F x = 0.  $$
Here it made no difference what vector $x$ might be, it is always zero.
Let $H$ be a symmetric matrix. What can you say about
$$  x^T (F+H)x?  $$
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Meanwhile, any square matrix can be written as the sum of a symmetric part and a skew symmetric part, in only one way. Given some $B$ we define the symmetric part as
$$  (B + B^T)/ 2, $$ the skew symmetric part as
$$  (B - B^T)/2 $$
A: The main idea is that many different matrices give rise to the same quadratic form, and out of these many matrices, we can always find a symmetric one.
Let's take an example of a quadratic form in two variables coming from a matrix:
$$\left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} 1 & 2 \\ 4 & 5 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = x^2 + 6xy + 5y^2$$
We can re-write this in matrix form as
$$\left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} 1 & 3 \\ 3 & 5 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = x^2 + 6xy + 5y^2$$
In this form, we have a symmetric matrix giving rise to $x^2 + 6xy + 5y^2$.
In general, the quadratic form
$$\left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = ax^2 + (b+c)xy + dy^2$$
can always be re-written in the form
$$\left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} a & (b+c)/2 \\ (b+c)/2 & d \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = a^2 + (b+c)xy + dy^2$$
A similar ideal works for quadratic forms in many variables.
