# 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.

• What does "written as" mean in this case? Change of base? Can you provide more detail?
– DRF
Commented Jan 15, 2016 at 20:57
• Well, the exact wording is "replace". Here's a link to the passage. i.sstatic.net/3upcw.jpg Commented Jan 15, 2016 at 21:02

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 $$(B - B^T)/2$$

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 $x^2 + 6xy + 5y^2$.

$$\left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = ax^2 + (b+c)xy + dy^2$$
$$\left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} a & (b+c)/2 \\ (b+c)/2 & d \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = a^2 + (b+c)xy + dy^2$$
• This idea is very nice. Could you please share the name of this theory? And how can we generalize it for squared matrices with size n x n Commented Mar 6, 2019 at 5:21
• It's not really a "theory", it's just an idea I came up with. In general, with a matrix $M$, the symmetric one would be $\frac{1}{2}(M+M^{\top})$. Commented Mar 6, 2019 at 16:58