The inner product space with $\langle f(x),g(x)\rangle=\int_0^2 f(t)g(t)dt$ on $P_1(\mathbb{R})$. I need to prove that $T(f(x))=f(x)-f(1)$ is an orthogonal projection.
I know that I need to show $R(T)^{\bot}=N(T)$ and $N(T)^{\bot}=R(T)$. I'm stuck on actually solving R(T) and N(T) given T(f(x))=f(x)-f(1).
I started with the basis of $P_1(\mathbb{R})$, so $R(T)=span(\{T(1),T(x)\})$, but I'm not sure what $T(1)$ and $T(x)$ equals.
If it helps, in a class example we had that $T(1)$ and $T(x)$ equals $span(\{1,0\})$ when $T(f(x))=f(0)$
Additionally, I began to set up the null, and ran into a similar problem. I have $N(T)=\{a+bx\in P_1\mathbb(R)|T(ax+b)\}$ so far.
For each $f(x)\in P_1(\Bbb R)$, $T\bigl(f(x)\bigr)$ is a polynomial such that $T\bigl(f(1)\bigr)=0$. This suggests that $T$ is the orthogonal projection on$$\{p(x)\in P_1(\Bbb R)\mid p(1)=0\}=\{ax-a\mid a\in\Bbb R\}.\tag1$$On the other hand $\ker T$ is the set of all constant polynomials (since $p(x)\in\ker T\iff p(x)-p(1)=0$). And, if $k$ is a constant polynomial, then$$\langle k,ax-a\rangle=\int_0^2ka(x-1)\,\mathrm dx=0.$$So, $T$ is indeed the orthogonal projection on $(1)$.