Does the set of skew-symmetric n×n matrices form a vector space? Does the set of skew-symmetric n×n matrices form a vector space with the usual matrix addition and scalar multiplication?
This is quite easy to prove if we take a specefic dimension like 2x2, but I am quite confused about poving it for all nxn square matricies.
 A: Let $\operatorname{Mat}_n$ be the set of $n \times n$ matrices and $\operatorname{Skew}_n = \left\{ A \in \operatorname{Mat}_n \;\middle|\; A^T = - A \right\}$ be the set of $n \times n$ skew-symmetric matrices. The answer to your question is yes. The easiest way to see this is by showing that $\operatorname{Skew}_n$ is a vector subspace of the vector space $\operatorname{Mat}_n$. To do that, let we have to show that 


*

*$0 \in \operatorname{Skew}_n$

*$A, B \in \operatorname{Skew}_n \implies A + B \in \operatorname{Skew}_n$

*$\alpha$ a scalar, $A \in \operatorname{Skew}_n \implies \alpha A \in \operatorname{Skew}_n$.


Here are the proofs:


*

*$0^T = 0 = -0$, so $0 \in \operatorname{Skew}_n$.

*$(A+B)^T = A^T + B^T = -A + (-B) = -(A+B)$

*$(\alpha A)^T = \alpha A^T = - \alpha A$.

A: Let $A$ and $B$ be $n\times n$ skew-symmetric matrices, with $(A)_{ij}=a_{ij}$ and $(B)_{ij}=b_{ij}$ such that $a_{ji} = -a_{ij}$, $b_{ji} =-b_{ij}$.
Let $C = A+ B$, so that $(C)_{ij} =a_{ij}+b_{ij}$. Now $(C)_{ji} = a_{ji} +b_{ji} =-(a_{ij}+b_{ij}) = (-C)_{ij}$ and so $C$ is skew-symmetric.
The last thing you need to check is that $\lambda A$ is skew symmetric for all scalars $\lambda$. Can you figure that out?
