What square matrices cannot be expressed as the sum of symmetric and skew-symmetric parts I thought I knew that any square matrix can be written as the sum of symmetric and antisymmetric matrices since we have the property that any $n\times n$ matrix $A$ can be expressed as $A=\frac{1}{2}(A-A^T)+\frac{1}{2}(A+A^T)$. However, I have come across the following statement and was told this is not always true and asked if I could find a counterexample where this isn't true but I have been struggling to find a counterexample (perhaps when $\mathbb{F}=\mathbb{Z}_2$?). Here is the statement:
Let $\mathbb{F}$ be a field. Then any $A\in \mathbb{F}(n,n)$ may be written as a sum of a symmetric and an anti-symmetric matrices in $\mathbb{F}(n,n)$.
I would greatly appreciate if someone could show me a counterexample to this.
 A: Another point, in addition to the failure of the standard formula, which is not instantly a proof of non-existence of the decomposition in characteristic $2$ (though close to it), is to observe that in char $2$ skew-symmetry is symmetry. So the question would be whether every matrix is a sum of symmetric and symmetric, hence symmetric. And we easily respond that, no, not every matrix in char 2 (of size bigger than $1\times 1$) is symmetric.
So, again, not only does the expected formula fail, ...
A: You have already made the key observation. The formula $A=\frac{1}{2}(A-A^T)+\frac{1}{2}(A+A^T)$ is not valid for a field of characteristic $2$. Look at the Galois field $F_4$ which has 4 elements $(0, 1, \alpha, 1+\alpha)$. The matrix $$\begin{pmatrix}0, 1 \\ \alpha, 0 \end{pmatrix}$$ has the desired form. Actually now that I think of it
$$\begin{pmatrix}0, 1 \\ 0, 0 \end{pmatrix}$$
is an example in $\mathbb{Z}_2$ as well.
A: Another way of understanding why this isn't always true, in lieu of a counterexample, is via a dimension argument.  $n\times n$ matrices may be viewed as living in a vector space $V = \mathbb F^{n\times n}$.  This space has dimension $n^2$.   (Use standard basis vectors of the form $\mathbf e_k\mathbf e_j^T$ and count.)
For fields with characteristic $\neq 2$ then we have two subspaces $S_y$ and $S_k$ consisting of symmetric and skew symmetric matrices respectively. By inspection:
$\dim\big(S_y\big) = \binom{n}{2} +n$
$\dim\big(S_k\big) = \binom{n}{2}$
so if
$V=S_y + S_k$ then $V=S_y \oplus S_k $
because there is a trivial intersection between $S_y$ and $S_k$
i.e. if we find some $X$ such that $X\in S_y$ and $X\in S_k$, then $X=X^T$ because it is symmetric, and $-X=X^T$ because it is skew symmetric.  So $X=-X$ or $2X=\mathbf 0\implies X=\mathbf 0$.
To confirm $V=S_y \oplus S_k $, check the dimensions.  i.e. if the RHS, a subspace of $V$, provides a basis with $n^2$ elements then it must be consist of all of $V$. To confirm:
$n^2 =\dim\big(V\big) =  \dim\big(S_y\big) + \dim\big(S_k\big)=\big(n + \binom{n}{2}\big) + \binom{n}{2}=n^2$
However in fields of characteristic 2 you have $S_k\subseteq S_y$,
if $V=S_y + S_k$ then  $V=S_y + S_k = S_y $  which further implies
$n^2= \dim\big(V\big)=\dim\big(S_y + S_k\big)=\dim\big(S_y \big) =\big(n + \binom{n}{2}\big) \lt n^2  $
which is impossible
note:
it is not necessarily true that $S_y\subseteq S_k$ in characteristic 2.  Many texts will explicitly define skew symmetry such that the matrix has all zeros on the diagonal hence $I_n$ is in $S_y$ but not $S_k$.  Conventions vary.  That said this convention is quite useful-- e.g. when working with skew symmetric bilinear forms and showing congruence of invertible skew symmetric matrices to the symplectic matrix over any field.
