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Let $V$ be a vector space ($\operatorname{dim}V<\infty$) over some field $\mathbb{F}$, $\operatorname{char}(\mathbb{F})\neq2$
and let $T:V\to V$ be linear such that $T\circ T=Id_V$.
Define $\quad U=\{v\in V|\:T(v)=v\}$,$\quad W=\{v\in V|\:T(v)=-v\}$

Show:
1) $V=U\oplus W$
2) $T=r_{U,W}$ (reflection)

Progress:
1) $$U\cap W=\{v\in V|\:T(v)=v\quad \operatorname{and}\quad T(v)=-v\}$$ The only vector which satisfies that is $0\in V$ therefore $U\cap W=\{0\}$.
My struggle here is with showing that $V=U+W$

Moreover, how does $\operatorname{char}(\mathbb{F})\neq2$ affect the problem?

Slavik Egorov
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2 Answers2

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Where does the characteristic being $\ne2$ enter the scene? Well, if the characteristic is $2$, then $U=W$.

If the characteristic is $\ne2$, then $U$ and $W$ are eigenspaces relative to $1$ and $-1$, so they're independent, that is, $U\cap W=\{0\}$ (but your direct proof is good as well).

You'd like to see that $U+W=V$. Given $v\in V$, you'd like to write $v=u+w$, with $T(u)=u$ and $T(w)=-w$. Suppose you have found them: then $T(v)=T(u)+T(w)=u-w$.

Hence we obtain $$ u=\frac{1}{2}(v+T(v)),\qquad w=\frac{1}{2}(v-T(v)) $$ It remains to prove $$ \frac{1}{2}(v+T(v))\in U,\qquad \frac{1}{2}(v-T(v))\in W $$ which is just a simple verification using that $T^2$ is the identity. Note that $1/2$ exists because the characteristic is $\ne2$.

Comment. This is essentially the same proof that every matrix over a field of characteristic $\ne2$ is uniquely the sum of a symmetric and an antisymmetric matrix (which is also a particular case).

egreg
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  • If I got it right, the characteristic being $2$ would mean that $2\equiv 0$ and from the properties of $U,W$ we can derive $v=-v\iff 2v=0$. How does that imply $U=W$? – Slavik Egorov Mar 30 '18 at 07:07
  • @SlavikEgorov If $v=-v$, then the definitions of $U$ and $W$ are the same. – egreg Mar 30 '18 at 07:39
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Another approach to solve 1) is to see that $v=(v-T(v))+T(v)$. It is easily seen that $v-T(v)\in W$ and $T(v)\in U$.

Notone
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