Show $M_n (\mathbb F) = W \oplus U $ and view a basis for $W$.

Question

Let $ M_n (\mathbb F) $ be the space of the matrices of order $ n $ with entries in the body $ \mathbb F $. Let $ W = \{ A \in M_n (\mathbb F): A + A^{t} = 0 \}$ and $U = \{ A \in M_n (F): A-A^{t} = 0 \}$. Show $M_n (\mathbb F) = W \oplus U $ and view a basis for $W$.

I tried to show that $ M_n (\mathbb F) \subset W \oplus U$ and $ W \oplus U \subset M_n (\mathbb F)$. For this I initially took $ A' \in M_n (F) $, however I lack ideas on how to prove that $A' \in W \oplus U$.

I also noticed that if $A \in W$, then

$$A = \left[ \begin{matrix} 2a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ -a_{1n} & \cdots & 2a_{nn} \\ \end{matrix} \right]$$

And if $B \in U$ then $$B = \left[ \begin{matrix} 0 & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{1n} & \cdots & 0 \\ \end{matrix} \right]$$

Does this help me with something to answer the question?

Answer 1

First question. $M_n (\mathbb F) = W \oplus U$ ($\text{If carac }\mathbb{F}\ne 2$).

(a) $$A\in W\cap U \Rightarrow A\in W\wedge A\in U\Rightarrow A^T=-A \;(*)\;\wedge A^T=A\;(**)$$ $$\underbrace{\Rightarrow}_{(**)-(*)} (1+1)A=0\underbrace{\Rightarrow}_{\text{If carac }\mathbb{F}\ne 2}A=0\Rightarrow W\cap U=\{0\}.$$ (b) If $A\in M_n(\mathbb{F})$ and denoting $1+1:=2$, $$A=\underbrace{\dfrac{1}{2}(A-A^T)}_{\in W}+\underbrace{\dfrac{1}{2}(A+A^T)}_{\in U}\Rightarrow M_n(\mathbb{F})=W+U.$$ Second question.

$$A\in W\Rightarrow A= \begin{bmatrix} 0 & -a_{21} & \ldots & -a_{n1}\\ a_{21} &0 & \ldots & -a_{n2} \\ \vdots&&&\vdots \\ a_{n1} & a_{n2} &\ldots & 0\end{bmatrix}=a_{21}\begin{bmatrix} 0 & -1 & \ldots & 0\\ 1 &0 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & 0\end{bmatrix}+\ldots +a_{n1}\begin{bmatrix} 0 & 0 & \ldots & -1\\ 0 &0 & \ldots & 0 \\ \vdots&&&\vdots \\ 1 & 0 &\ldots & 0\end{bmatrix}$$ $$+a_{n2}\begin{bmatrix} 0 & 0 & \ldots & 0\\ 0 &0 & \ldots &- 1 \\ \vdots&&&\vdots \\ 0 & 1 &\ldots & 0\end{bmatrix}+\ldots +a_{n,n-1}\begin{bmatrix} 0 & \ldots & 0 & 0\\ \vdots&&&\vdots \\ 0 & \ldots & 0 & -1\\ 0 & \ldots & 1 & 0\end{bmatrix}.$$ Those matrices are generators of $W$ and linearly independent, so $$\dim W= (n-1)+(n-2)+\ldots+2+1=\displaystyle\frac{n(n-1)}{2}=\displaystyle\binom{n}{2}.$$

Answer 2

There is one thing to note: the characteristic of $\mathbb{F}$ is not $2$. This is clear because the identity matrix is in both subspaces if the characteristic is $2$ (and that is not possible in a direct sum). Why do $A$ and $B$ take the forms you described? It seems like $B$ can have any value on the diagonal (including $0$). Maybe $A$ should have $0$ on the diagonal.

I have the following:

For $A\in M_n(\mathbb{F})$ let $A_w=\frac{A-A^T}{2}$ and $A_v=\frac{A+A^T}{2}$. Then $A_w\in W$ and $A_v\in V$. This gives us the hard inclusion: $M_n(\mathbb{F})\subset W\oplus V$.

Note: I did not show that the sum was direct