Let $W_1$ be the subspace of $\mathcal{M}_{n \times n}$ that consists of all $n \times n$ skew-symmetric matrices with entries from $\mathbb{F}$, and let $W_2$ be the subspace of $\mathcal{M}_{n \times n}$ consisting of all symmetric $n \times n$ matrices. Prove that $\mathcal{M}_{n \times n}(\mathbb{F}) = W_1 \oplus W_2$.

I couldn't really figure out why the sum of an $n \times n$ symmetric matrix and $n \times n$ skew-symmetric matrix would form a $n \times n$ matrix (to satisfy the direct summand property $W_1 + W_2 = \mathcal{M}_{n \times n}(\mathbb{F})).$ Browsing online, I found that $$ M = \frac{1}{2}(M + M^{t}) + \frac{1}{2}(M-M^{t}),$$ where $\frac{1}{2}(M+M^{t}) \in W_2, \frac{1}{2}(M-M^t) \in W_1$, and $M \in \mathcal{M}_{n \times n}(\mathbb{F})$.

This reminds me of a formula on how odd and even functions may be be added together to form a generic function. $$f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}.$$

However, to me it is disturbing to use unless I know how it was derived. If anyone could show me why a skew-symmetric matrix may be represented as $\frac{1}{2}(M-M^t)$ and why a symmetric matrix may be represented as $\frac{1}{2}(M+M^t)$ than I may sleep better tonight.



Rather than asking why a symmetric matrix may be represented as $\frac{1}{2}(M+M^t)$, you need to ask yourself the following three things:

  1. Is $\frac{1}{2}(M+M^t)$ symmetric, for every $M$?
  2. Is $\frac{1}{2}(M-M^t)$ skew-symmetric, for every $M$?
  3. Is the sum of these equal to $M$?

After you have answered "yes" to the above three questions, you will have proven that $M_{n,n}(\mathbb{F})=W_1+W_2$. Then, you merely prove that $W_1\cap W_2=\{0\}$, and then you have $M_{n,n}(\mathbb{F})=W_1\oplus W_2$.

Ultimately, there are many matrices $M$ that all have the same symmetrization via the above formula. The reason those two formulas are used is that they make the above three questions all have the right answer.

  • $\begingroup$ I see. Maybe I was overcomplicating this then. Good answer. $\endgroup$ – St Vincent Jan 14 '14 at 18:12
  • $\begingroup$ How to show $ W_1∩W_2=\{0\}$ $\endgroup$ – Aaron Abbasi Oct 18 '15 at 18:12
  • $\begingroup$ @ClairSymmetry Suppose a matrix is both symmetric and skew-symmetric. Then $A = A^t = -A^t = -A$ and so $A = 0$. $\endgroup$ – Sir Jective Feb 19 '16 at 5:13

There is nothing mysterious about these decompositions. Any linear operation $T$ on a vector space over a field$~F$ not of characteristic$~2$ that is an involution ($T^2=I$) is diagonalisable with eigenvalues $1,-1$ only, and the space is therefore the direct sum of the eigenspaces for $1$ and for $-1$. One may take $T$ for instance to be transposition of matrices, or argument reversal $f\mapsto(x\mapsto f(-x))$ in a vector space of functions. These eigenspaces are the cases of symmetric and anti-symmetric objects.

In fact in situations like this (operators satisfying a polynomial equation, where the polynomial splits into distinct linear factors corresponding to the potential eigenspaces) the projections on the eigenspaces can be expressed as polynomials in $T$. In the case at hand this means one can express symmetrising and anti-symmetrising operations for $T$ in terms of $T$ itself, and that is what the expressions giving components in $W_1$ and $W_2$ do. While in general such expression can be complicated, they are quite simple for involutions: the projection on the eigenspace for $1$ is $P_+=\frac12(T+I)$ and the projection on the eigenspace for $-1$ is $P_-=\frac12(-T+I)$. One easily checks $T\cdot P_+=P_+$ and $T\cdot P_-=-P_-$, confirming that on the images of the projection $T$ acts as$~1$ respectively as$~{-}1$. Of course one also has $P_++P_-=I$, so that a value (matrix) can be reconstructed from its symmetrised and anti-symmetrised parts. The fact that all this is quite similar to decomposing a function into an even and an odd function is therefore no coincidence.

  • $\begingroup$ Thanks for the addition here. $\endgroup$ – St Vincent Jan 24 '14 at 1:55

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