what is the matrix of linear transformation of $T(w_1 + w_2) = w_1 - w_2$ $\forall$ $w_1 \in W_1$, $w_2 \in W_2$

Question

Let $V$ be a vector space of all real valued functions from $(-1,1) \rightarrow \mathbb{R}$, $W_1 = \{f \in V \vert f(x) = f(-x)$ $\forall x \in (0,1)\}$, $W_2 = \{f \in V \vert f(-x) = -f(x)$ $\forall x \in (0,1)\}$. Let $T : V \rightarrow V$be a linear operator such that $T(w_1 + w_2) = w_1 - w_2$ $\forall$ $w_1 \in W_1$, $w_2 \in W_2$, then

1. $T$ is invertible
2. $T$ is involutory
3. $Ker(T-I) \cap Ker(T+I) = \{0\}$
4. $Im(T-I) \cap Im(T+I) = \{0\}$

I am unable to figure out how to proceed with this question. What would the matrix of the transformation look like? Can someone help me solve this question?

Answer 1

First of all we have $V=W_1 \oplus W_2$. Are you able to show this ?

Furthermore: $T^2(w_1+w_2)=T(w_1-w_2)=w_1-(-w_2)=w_1+w_2$ for all $w_1 \in W_1,w_2 \in W_2$.

Hence $T^2=id_V$. This shows 2. and 1.

If $v \in Ker(T-I) \cap Ker(T+I)$, then $Tv=v$ and $Tv=-v$, hence $v=-v$, therefore $v=0$. This gives 3.

Now it is your turn to show 4.