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Let $V$ be a vector space over $\mathbb{C}$ such that $T^2=1_V$. Define

$V_1 = \left\{v\in V |\ T(v) = v \right\}$, $\ V_2 = \left\{v\in V |\ T(v) = -v\right\}$

prove that $V = V_1 \oplus V_2$

For start, I've shown that $V_1\ ,V2$ are subspace of $V$. However, I'm stuck trying to show that $\forall \ v\in V,\ v = v_1 + v_2$ where $v_1 \in V1, v_2 \in V_2$.

I know at some point I have to make use of $T^2=1_V$, but cant figure out how. Any help or insight is deeply appreciated.

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Hint: To prove that $V=V_1+V_2$, we have to prove that for each $v\in V$ there are $v_1\in V_1$ and $v_2\in V_2$ such $v=v_1+v_2$. So, take $v\in V$ and consider $$ v_1 = \dfrac12 (v+T(v)) \quad v_2 = \dfrac12 (v-T(v)) $$

Then $v_1\in V_1$, $v_2\in V_2$, and $v=v_1+v_2$.

It remains to prove that $V_1 \cap V_2 = 0$:

If $v \in V_1 \cap V_2$, then $v=T(v)=-v$ and so $2v=0$.

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  • $\begingroup$ I think I understand what you are trying to do here, but I have one concern is that we are kinda assuming $\forall v_1\in V_1$, is of the form $v_1 = \frac{1}{2}(v + T(v))$ and vice versa for $v_2$. why can we do that in this case ? $\endgroup$ – some1fromhell Apr 14 '16 at 11:21
  • $\begingroup$ @user2875613, we just have to exhibit $v_1$ and $v_2$ given $v$. $\endgroup$ – lhf Apr 14 '16 at 11:23
  • $\begingroup$ I see , so the point is we want to show $\forall v\in V$, $v$ can written as SOME sum of 2 vectors from $V_1, V_2$ right ? $\endgroup$ – some1fromhell Apr 14 '16 at 11:26
  • $\begingroup$ @user2875613, yes. $\endgroup$ – lhf Apr 14 '16 at 11:36
  • $\begingroup$ I see, thank you for the reply. It really helped my revision $\endgroup$ – some1fromhell Apr 14 '16 at 11:37

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