Exercise Let $V$ be any vector space with $T:V \to V$. Suppose $M_1 =\{v \in V: T(v) = v\}$, $M_2 = \{v \in V: T(v) = -v\}$ are eigenspaces of $V$ and $T^2(v) = v$, show that $V = M_1 + M_2$, i.e. show that
$$\forall \, v \in V, v = m_1+m_2 \hspace{0.3cm} m_1 \in M_1, m_2 \in M_2$$
So $v$ can be written as the sum of objects from $M_1$ and $M_2$.
Attempt.
Let $v \in V$. We have that $T^2(v) = v$.
Perhaps we could consider $v = v+0 = v +(v+(-v))$. Then
$$T(v) = T(v) +T(v) + T(-v) = 2T(v) - T(v)$$
I'm not sure how to utilize $T^2(v) = v$ here. How can we show that $v$ is the sum of two objects from $M_1$ and $M_2$ respectively when $v$ is chosen arbitrarily?
As @paul garrett hinted :
The element $\frac{v+T(v)}{2} \in V $ and also $\frac{v+T(v)}{2} \in M_1 $ because $T(\frac{v+T(v)}{2} ) = \frac{v+T(v)}{2} $ given that $T^2(v) = v$.
$\frac{v-T(v)}{2} \in M_2$ for the same reason.
For any $v \in V$ then these choices are fine.