Find kernel and image of the polynomial derivative and its adjugate.

Question

I have the mapping $f: \mathbb{R}[t]_{\leq n} \to \mathbb{R}[t]_{\leq n}$;

$p = \sum_{i=1}^{n} \alpha_{i}t^{i} \mapsto \sum_{i=1}^{n} i\alpha_{i}t^{i-1} = p'(t)$

equipped with scalar product

$\langle p, q\rangle = \int_{-1}^{1} p(t)*q(t)\,dt$

To find the adjugate $f^*$, I do the following:

Let $p = \sum_{i=1}^{n} \alpha_{i}t^{i}$ and $q = \sum_{j=1}^{m} \beta_{j}t^{j}$.

$\langle f(p), q\rangle = \int_{-1}^{1} p'(t)*q(t)\,dt = p(1)q(1) - p(-1)q(-1) - \int_{-1}^{1} p(t)*q'(t)\,dt$

$ = (\sum_{i=1}^{n} \alpha_{i}(1)^{i}\sum_{j=1}^{m} \beta_{j}(1)^{j}) - (\sum_{i=1}^{n} \alpha_{i}(-1)^{i}\sum_{j=1}^{m} \beta_{j}(-1)^{j}) - \langle p, f(q)\rangle$

$= (\sum_{i=1}^{n}\sum_{j=1}^{m} \alpha_{i}\beta_{j}(1)^{i+j}) - (\sum_{i=1}^{n}\sum_{j=1}^{m} \alpha_{i}\beta_{j}(-1)^{i+j}) - \langle p, f(q)\rangle$

$= (\sum_{i=1}^{n}\sum_{j=1}^{m} \alpha_{i}\beta_{j}(1 - (-1)^{i+j}) - \langle p, f(q)\rangle$

Beyond this I am having trouble determining how exactly to define $f^*$.

Furthermore, any hints on how to find the kernel of $f^*$, the image of $f$, and its orthogonal complement are appreciated, since I am very, very new at this.