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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.

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    $\begingroup$ Yes it does and the dimension is $n(n-1)/2$ $\endgroup$ – user379195 Mar 19 '17 at 14:25
  • $\begingroup$ They form a vector space. Are you looking for a proof? $\endgroup$ – Rafael Wagner Mar 19 '17 at 14:25
  • $\begingroup$ @RafaelWagner yes look for a proof $\endgroup$ – Erik Hambardzumyan Mar 19 '17 at 14:26
  • $\begingroup$ See here they write 'Sums and scalar multiples of skew-symmetric matrices are again skew-symmetric. Hence, the skew-symmetric matrices form a vector space. Its dimension is $n(n−1)/2$.' $\endgroup$ – Rafael Wagner Mar 19 '17 at 14:26
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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

  1. $0 \in \operatorname{Skew}_n$
  2. $A, B \in \operatorname{Skew}_n \implies A + B \in \operatorname{Skew}_n$
  3. $\alpha$ a scalar, $A \in \operatorname{Skew}_n \implies \alpha A \in \operatorname{Skew}_n$.

Here are the proofs:

  1. $0^T = 0 = -0$, so $0 \in \operatorname{Skew}_n$.
  2. $(A+B)^T = A^T + B^T = -A + (-B) = -(A+B)$
  3. $(\alpha A)^T = \alpha A^T = - \alpha A$.
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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?

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