The linear operator $L:V\to V$ is defined by $L(p(x)) = -6p'(x) - 6p(x)$. How can I find the adjoint $L^{*}$ of $L$? If I'm given an inner produce:
$$\langle p(x),q(x)\rangle = \int_{0}^{1}p(x)q(x)\mathrm{d}x$$
on the vector space $V$ of real polynomials with degree less than $2$.
The linear operator $L:V\to V$ is defined by $L(p(x)) = -6p'(x) - 6p(x)$.
How can I find the adjoint $L^{*}$ of $L$.
I'm not really sure how to start. Any tips?
 A: In all cases, we start from the definition of $L^*$: it's the unique linear operator such that
$$
\langle Lp,q \rangle = \langle p, L^*q \rangle.
$$
Whenever a derivative operator is involved, integration by parts is a way to transfer the derivative to the other side of the product:
$$
\langle p'(x), q(x) \rangle = - \langle p(x), q'(x) \rangle + p(1) q(1) - p(0)q(0).
$$
In particular, for $Lp = -6p' - 6p$, we have
$$
\langle Lp, q \rangle = \langle p, 6q' - 6q \rangle -6\left[p(1) q(1) - p(0) q(0)\right].
$$
Therefore it remains to compute which linear operator $H$ on $V$ corresponds to the equation
$$
\langle p, Hq\rangle = p(1) q(1) - p(0) q(0),
$$
after which the adjoint will be defined by $L^*p = 6p' - 6p - 6Hp$.
The space $V$ is specified here as the space of polynomials of degree less than $2$. In particular, it is a two dimension space with an orthogonal basis is given by $\{p_0, p_1\}$, where $p_0(x) = 1$ and $p_1(x) = \sqrt{12} (x - 1/2)$. You might find it worthwhile to check that this is indeed an orthogonal basis, and also check that you can construct it by applying Gram-Schmidt on the usual polynomial basis $\{1, x\}$.
Then we compute
$$
\langle Hp_0, p_0\rangle = 0, \quad \langle Hp_0, p_1 \rangle = \langle Hp_1,  p_0 \rangle = \sqrt{12}, \quad \langle Hp_1, p_1 \rangle = 0.
$$
In particular, this means that
$$
Hp_0 = H(1) = \sqrt{12} p_1 = 12(x - 1/2),
$$
and
$$
Hp_1 = \sqrt{12}p_0 = \sqrt{12}.
$$
Since we have specified how $H$ acts on a basis of $V$, we have in essence computed it, and as stated before, the adjoint of $L$ is defined by $L^* p = 6p' - 6p - 6Hp$ for $p \in V$.
You might find it worthwhile to explicitly work out a formula for $H$ if it helps you understand how to construct a linear operator based on its matrix representation.
