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 $L$?

Question

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?

Answer 1

There's no really simple way to compute adjoints. It tends to be nicer to compute the adjoint on vectors in a basis, rather than deal with the whole thing at once. So, let's compute the adjoint of $x^2$. Using the definition of adjoint, $$\langle T(p), x^2 \rangle = \langle p, T^*(x^2)\rangle,$$ for all polynomials $p$.

Note: the unknown here is $T^*(x^2)$. We should be able to identify a unique vector $T^*(x^2)$ by considering $p(x) = 1$, then $p(x) = x$, then $p(x) = x^2$ (or any other basis). Let $T^*(x^2) = ax^2 + bx + c$.

If $p(x) = 1$, then we get the equation \begin{align*} &\langle T(1), x^2 \rangle = \langle 1, T^*(x^2)\rangle \\ \implies &\int_0^1 -6x^2 \, \mathrm{d}x = \int_0^11 \cdot (ax^2 +bx + c) \, \mathrm{d}x \\ \implies &-2 = \frac{a}{3} + \frac{b}{2} + c. \end{align*} If $p(x) = x$, then \begin{align*} &\langle T(x), x^2 \rangle = \langle x, T^*(x^2)\rangle \\ \implies &\int_0^1 (-6x - 6)x^2 \, \mathrm{d}x = \int_0^1x \cdot (ax^2 +bx + c) \, \mathrm{d}x \\ \implies &-\frac{7}{2} = \frac{a}{4} + \frac{b}{3} + \frac{c}{2}. \end{align*} If $p(x) = x^2$, then \begin{align*} &\langle T(x^2), x^2 \rangle = \langle x^2, T^*(x^2)\rangle \\ \implies &\int_0^1 (-6x^2 - 12x)x^2 \, \mathrm{d}x = \int_0^1x^2 \cdot (ax^2 +bx + c) \, \mathrm{d}x \\ \implies &-\frac{21}{5} = \frac{a}{5} + \frac{b}{4} + \frac{c}{3}. \end{align*} Now we just solve the system: $$\left\lbrace \begin{aligned} \frac{a}{3} + \frac{b}{2} + c &= -2 \\ \frac{a}{4} + \frac{b}{3} + \frac{c}{2} &= -\frac{7}{2} \\ \frac{a}{5} + \frac{b}{4} + \frac{c}{3} &= -\frac{21}{5}, \end{aligned} \right.$$ or the matrix equation $$\begin{pmatrix} \frac{1}{3} & \frac{1}{2} & 1 \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\ \frac{1}{5} & \frac{1}{4} & \frac{1}{3} \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} -2 \\ -\frac{7}{2} \\ -\frac{21}{5} \end{pmatrix}. $$

I strongly recommend finding the inverse to this $3 \times 3$ matrix, as you will be solving a system with the same coefficient matrix two more times (one for $x$ and one for $1$). Computing the inverse, we get $$\begin{pmatrix} \frac{1}{3} & \frac{1}{2} & 1 \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\ \frac{1}{5} & \frac{1}{4} & \frac{1}{3} \end{pmatrix}^{-1} = \begin{pmatrix} 30 & -180 & 180 \\ -36 & 192 & -180 \\ 9 & -36 & 30 \end{pmatrix}.$$ So, $$\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 30 & -180 & 180 \\ -36 & 192 & -180 \\ 9 & -36 & 30 \end{pmatrix} \begin{pmatrix} -2 \\ -\frac{7}{2} \\ -\frac{21}{5} \end{pmatrix} = \begin{pmatrix} -186 \\ 156 \\ -18 \end{pmatrix}. $$ So, all in all, we can conclude, $$T^*(x^2) = -186x^2 + 156x - 18.$$ Next, you can compute $T^*(x)$ and $T^*(1)$ in the same way (and the inverse computed above should be useful once again). Then, $$T^*(ax^2 + bx + c) = a(-186x^2 + 156x - 18) + bT^*(x) + cT(1),$$ whatever these polynomials happen to be.

---

The alternative is to use CyclotomicField's suggestion and turn it into a matrix. This means you need to:

1. Find an orthonormal basis. This usually involves computing Gram-Schmidt on a basis like $(1, x, x^2)$. This is already a long step!
2. Compute the matrix with respect to the basis. Fortunately, when you have an orthonormal basis, this is quite straightforward, as the entries of the matrix will just be inner products (though this still requires some computation).
3. Transpose the matrix (trivial).
4. Write out an arbitrary polynomial $ax^2 + bx + c$ as a coordinate column vector with respect to the orthonormal basis. This is quick and easy, as the basis is orthonormal.
5. Compute the product of the transposed matrix with the coordinate column vector (some computation, but easy enough).
6. Convert this coordinate column vector back to a polynomial (easy).

In total, I'm not convinced it's much quicker!

Answer 2

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.