Show that 2 matrices belong to a square matrix by taking the transpose. Vector spaces

Question

Let $M_n(K)$ be an $n\times n$ matrix of a K vector space.

\begin{align} U &= \{A ∈ M_n(K) | A_{ij} = A_{ji} \} \\ W &= \{A ∈ M_n(K) | A_{ij} = -A_{ji} \} \end{align}

Prove that $U$ and $W$ are subspaces of $M_n(K)$

The mark scheme shows that its closed under addition and multiplication by a scalar. but then my friend said an alternative is to take the transpose. where $(A+B)^T = A^T + B^T = A+B$. How does that work? Is it because it's a square matrix?

Also the second part of the question was to prove that $M_n(K)$ does not equal $U ⊕ W$ when $K$ is the field $F_2$. ?

Answer 1

Note that $U := \{A \in M_n(K): A^T = A\}$ and $W := \{A \in M_n(K): -A^T = A\}$. So the idea of your friends is the same you had (checking directly that it is closed under multiplication and addition) but you two write it differently.

Now, if $K = \mathbb F_2$, note that the matrix $I\in M_n(K)$ with $I_{i,j} = 1$ for every $1 \leq i,j \leq n$, is such that $I^T = I$ and thus $I \in U$ but we also have $-I^T = -I = I$ (because $-1 = 1$ in $\mathbb F_2$) and so $I \in W$. It follows that $W \cap U \neq \{0\}$ and so the sum can't be direct.

Answer 2

Here's a hint at an alternative proof that $U$ and $W$ are subspaces of $M_n(K)$.

Let $T_U,T_W\colon M_n(K)\to M_n(K)$ be given by $T_U(A)=A-A^T$ and similarly $T_W(A)=A+A^T$. Show that these are linear transformations. Show that $\ker T_U=U$ and $\ker T_W=W$. Deduce that $U$ and $W$ are subspaces of $M_n(K)$.