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I have a very simple question Let $V$ be the real (finite-dimensional) inner product space of polynomials of degree at most $2$, with inner product $(p,q):= \int_0^1 p(x)q(x) \, dx$.

Let $T$ be the linear operator that maps $p_0+p_1x+p_2x^2$ to $p_1x$. The question is how to find the adjoint of $T$.

I have one way to solve the problem: namely find an orthonormal basis of $V$, find the matrix associated to $T$ under this basis and then take its transpose to get the matrix of $T^*$ under this basis.

Is there any quicker/smarter way?

Thanks a lot in advance

Jenny

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It is funny to note that integration by parts

$$(Tp,q)=\int_0^{1}p_1 x q(x)dx=p_1\left[g(1)-\int_{0}^1 g(x)dx \right],$$

provides us with little information as $g(x)$, which satisfies $g'(x)=q(x)$ is not an element of $V$ if the coefficient $q_2$ of the polynomial

$$q(x)=q_0+q_1x +q_2x^2$$

is different from $0$.

Let us try another method. By definition, $T^{\dagger}(q)\in V$, where

$$T^{\dagger}(q):=\alpha_0+\alpha_1 x+\alpha_2 x^2,$$

with $\alpha_i=\alpha_i(q_0,q_1,q_2)$ for all $i$ and

$$(T(p),q)=(p,T^{\dagger}(q)),$$

for all $p,q\in V$. We select different $p$'s and try to arrive at information for the functions $\alpha_i$.

  • $p(x)=1$

In this case we have $0=(T(1),q)=(1,T^{\dagger}(q))$, or

$$\int_0^{1} T^{\dagger}(q)dx=\alpha_0+\frac{\alpha_1}{2}+\frac{\alpha_2}{3}=0.$$

  • $p(x)=x$

In this case we have $(T(x),q)=(x,T^{\dagger}(q))$, or

$$\int_0^{1} x q(x) dx=\int_0^{1} xT^{\dagger}(q)dx$$

which is equivalent to $\delta=\frac{\alpha_0}{2}+\frac{\alpha_1}{3}+\frac{\alpha_2}{4},$ where

$$\delta=\frac{q_0}{2}+\frac{q_1}{3}+\frac{q_2}{4}. $$

  • $p(x)=x^2$

In this case we have $0=(T(x^2),q)=(x^2,T^{\dagger}(q))$, or

$$\int_0^{1} x^2T^{\dagger}(q)dx=\frac{\alpha_0}{3}+\frac{\alpha_1}{4}+\frac{\alpha_2}{5}=0.$$

In summary, we arrive at the system of equations

$$\alpha_0+\frac{\alpha_1}{2}+\frac{\alpha_2}{3}=0 $$ $$\frac{\alpha_0}{2}+\frac{\alpha_1}{3}+\frac{\alpha_2}{4}=\delta $$ $$\frac{\alpha_0}{3}+\frac{\alpha_1}{4}+\frac{\alpha_2}{5}=0 $$

which admits the solution

$$\alpha_0=-36\delta $$ $$\alpha_1=192\delta $$ $$\alpha_2=-180\delta. $$

I do not know if this method is quicker than orthonormalization, but I hope it helps.

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  • $\begingroup$ I dont know if it s quicker either.. but it was very useful to me for sure!! Thank you! $\endgroup$ – Jenny Jun 30 '13 at 20:28

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