Exercise 1.3.28

Question

A matrix $M$ is called skew-symmetric if $M^{t}=-M$. Clearly, a skewsymmetric matrix is square. Let $F$ be a field. Prove that the set $\mathrm{W}_{1}$ of all skew-symmetric $n 	imes n$ matrices with entries from $F$ is a subspace of $\mathrm{M}_{n 	imes n}(F)$. Now assume that $F$ is not of characteristic two (see page 549), and let $\mathrm{W}_{2}$ be the subspace of $\mathrm{M}_{n 	imes n}(F)$ consisting of all symmetric $n 	imes n$ matrices. Prove that $\mathrm{M}_{n 	imes n}(F)=\mathrm{W}_{1} \oplus \mathrm{W}_{2}$.

Jephian C.-H. Lin · Accepted Answer

\begin{proof}
By the previous exercise, we have $(M_1+M_2)^t=M_1^t+M_2^t=-(M_1+M_2)$ and $(cM)^t=cM^t=-cM$. With addition that zero matrix is skew-symmetric we have the set of all skew-symmetric matrices is a space. We have $$M_{n	imes n}(\mathbb{F})=\{A:A \in M_{n	imes n}(\mathbb{F})\}=\{(A+A^t)+(A-A^t):A \in M_{n	imes n}(\mathbb{F})\}=W_1+ W_2$$ and $W_1\cap W_2=\{0\}$. The final equality is because $A+A^t$ is symmetric and $A-A^t$ is skew-symmetric. If $\mathbb{F}$ is of characteristic 2, we have $W_1=W_2$.
\end{proof}

Exercise 1.3.28

Answers

Comments

Exercise 1.3.28

Answers

Comments

Add answer