VERIFYING: Proving an $n \times n$ Matrix Vector Space

Question

Let $V$ be the vector space made by the $n \times n$ square matrices.

a) Prove that $S=\{A\in V|A^t=A\}$ is a subspace of $V$

b) Prove that $T=\{A\in V|A^t=-A\}$ is a subspace of $V$

c) Prove that $V=S\oplus T$

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SOLUTION

a) Let $O$ be the zero matrix, then $O \in S$ because $O^t=O$.

Then, let $A, B \in S$; and $r, k \in \mathbb R$. Hence

$$rA+kB=r(A^t)+k(B^t)=rA^t+kB^t$$ then $rA+kB \in S \implies$ $S$ is a subspace of $V$

b) Let $O$ be the zero matrix, then $O \in T$ because $O^t=O$.

Then, let $A, B \in T$; and $r, k \in \mathbb R$. Hence

$$rA+kB=r(-A^t)+k(-B^t)=-rA^t-kB^t=-(rA^t+kB^t)$$ then $rA+kB \in T \implies$ $T$ is a subspace of $V$

c) I have no idea how to do it.

I HOPE YOU CAN HELP ME VERYFING THE a), b) parts and with the part c).

Answer 1

Actually, instead of taking a full linear combination, you can consider only $rA + B$, the $r$ scalar will guarantee the closure over the operations. In part a), $O \in S$ you did ok. But, we want to check that $(rA + B)^t = rA + B$, you skipped a step there. Doing all the steps, we have: $$(rA + B)^t = (rA)^t + B^t = rA^t + B^t = rA + B$$ therefore $rA + B \in S$. For item b), just to be more careful, we should write $O^t = - O$, but that is just a detail. For the closure, I think you skipped the same step as before. Doing all the steps: $$(rA + B)^t = (rA)^t + B^t = rA^t + B^t = r(-A) + (-B) = - (rA+B)$$ and so, $rA + B \in T$.

For item c), just notice that you can always write $A = \dfrac{A + A^t}{2} + \dfrac{A - A^t}{2}$. Can you check that the first part is symmetric, and the second is anti-symmetric?

Answer 2

Your proofs for a) and b) look fine.

For c) any matrix $A \in V$, we can write $A = \dfrac{1}{2}(A+A^T) + \dfrac{1}{2}(A-A^T)$.

It is easy to see that $\dfrac{1}{2}(A+A^T)$ is symmetric while $\dfrac{1}{2}(A-A^T)$ is anti-symmetric.

Hence, $\dfrac{1}{2}(A+A^T) \in S$ and $\dfrac{1}{2}(A-A^T) \in T$. Therefore, $A \in S \oplus T$.

Since $A \in V \implies A \in S \oplus T$, we know that $V \subseteq S \oplus T$.

Showing $S \oplus T \subseteq V$ is easy. Therefore, $V = S \oplus T$.