Intuition for why any square matrix is a sum of a symmetric matrix and a skew-symmetric matrix?

Question

I've seen proofs for this theorem and understood then. However, I feel like I can gain a better intuition.

1. I would never think of this theorem myself. How could someone discover this theorem?

2. Is there a visual intuition?

3. Any other things that you think can help me gain a better intuition for that theorem?

(Feel free to answer even without answering all of the above questions. Such answers can still be helpful :) )

Answer 1

This is a nice question I had at one point, too. I'll echo the explanation I got at the time.

If $V$ is a vector space, a linear map $P\colon V\to V$ satisfying $P^2 = P$ is called a "projection." Given any projection, it is always possible to write a vector in a unique way as a sum of something in the kernel and something in the image. For matrices and the symmetric + skew-symmetric decomposition, you can check that $$ P(A) = \frac12(A-A^T) $$ is a projection whose kernel is the symmetric matrices and whose image is the skew symmetric matrices. In the case of functions on $\mathbb R$ and the even + odd decomposition, you can prove $$ P(f)(x) = \frac12(f(x)-f(-x)) $$ is a projection whose kernel is the even functions and whose image is the odd functions.

Even more, these two specific projections both arise in the following way. An involution is a linear map $T\colon V\to V$ such that $T^2 = I$, where $I\colon V\to V$ is the identity map $I(v) = v$. If $T$ is an involution, then $\frac12(I-T)$ is a projection. The involution in the case of matrices is $T(A) = A^T$ (take the transpose), and the involution in the case of functions is the reflection map $T(f)(x) = f(-x)$.

Another involution you may like to think about is the complex conjugation map that takes a complex number $z = x + iy$ to the complex number $\bar z$ defined by $$ \bar z = x - iy. $$ Then we can also realize the decomposition $$ z = \mathrm{re}(z) + i\mathrm{im}(z) = (\frac{z+\bar z}{2}) + i(\frac{z-\bar z}2) $$ as an example of this phenomenon.

Answer 2

Some stuff, sometimes, appear by magic, I remember when I was trying to play with matrices once, I wrote $B = A + A^t$, and I said hmm, isn't this symmetric? Will yeah $B$ is in fact symmetric since $B^t = (A+ A ^t)^t = A^t + (A^t)^t = A^t + A = B$, what about if I chose $C = A - A^t$ is it symmetric too or maybe anti-symmetric? Redoing the same steps we observe that $C^t = -C$.

Now let's just remember we didn't mention anything regarding $A$, we didn't put any restriction on it (of course it has to be a square matrix), but the fact that

\begin{cases} B^t = (A + A^t)^t = B\\ C^t = (A - A^t)^t = -C \end{cases}

Is interesting since if we add $B$ to $C$ it would yield $$ B + C = A + A^t + (A - A^t) = 2A = B^t - C^t \implies A = \frac{1}{2}\left(B^t - C^t\right) $$

Which is really cool

Answer 3

The intuition really is as follows: try to solve a $2\times 2$ system of linear equations!

Let's say you have a matrix $A$ and you want to write it as a sum of a symmetric matrix $X$ and a skew-symmetric matrix $Y$, i.e. you have the problem "Let $A=X+Y$, solve in $X,Y$".

Now, the only thing you know about $X$ and $Y$ is their behaviour around transpose, so you transpose the previous equation (what else can you do?), and you get $A^\top=(X+Y)^\top=X^\top+Y^\top=X-Y$

Right, so now you have a system of two equations with two unknowns:

$$\begin{array}{rcl}X+Y&=&A\\X-Y&=&A^\top\end{array}$$

which you can then solve using your favourite method (Gaussian elimination or whatever else you can think of) to get $X=\frac{1}{2}(A+A^\top)$, $Y=\frac{1}{2}(A-A^\top)$.

Answer 4

Let us consider the set $S$ of symmetric matrices.

The answer of Alex Ortiz is in terms of projection :

$$s(A):=\frac12(A+A^T)$$

is the projection of $A$ onto $S$.

As you ask about intuition, there is a cousin way to arrive at the same concept : $s(A)$ as defined by (1) is the closest symmetric matrix to $A$ closest in terms of a matrix norm. Any norm could be taken (indeed, all matrix norms are equivalent). I choose here to use the spectral norm.

Proof :

Let $C$ be a generic element of $S$ (we have therefore $C=C^T$).

$$A-s(A)=A-\frac12(A+A^T)=\frac12(A-A^T)=\frac12((A-C)-(A-C)^T)$$

As a consequence :

$$\|A-s(A)\|\leq\frac12(\|A-C\|+\|(A-C)^T\|)=\|A-C\|$$

because the norm of a matrix is equal to the norm of its transposed matrix (see for example here a proof for the spectral norm : Squared norm of matrix equal to squared norm of its transpose).

A same kind of reasoning can be made for the closest antisymmetric matrix to a given matrix $A$.

In fact we could introduce the dot product $<A,B>=tr(A^TB)$ as here and prove that the two spaces, the space of symmetric matrices and the space of antisymmetric matrices are orthogonal.