Find an orthonormal base for the subspace $S=\left\{p(t)\in P_3(\mathbb{R}): tp'(t)=p(t)\right\}$ and find also a base for the space $S^⊥$.

Question

Find an ortonormal base for the subspace $S=\left\{p(t)\in P_3(\mathbb{R}): tp'(t)=p(t)\right\}$

where $$\langle p,q\rangle=\int_{0}^{1}p(t)q(t)dt$$

And find also a base for the space $S^⊥$.

My attempt:

The only polynomials that satisfy $tp'(t)=p(t)$ are $p(t)=ct$ where $c\in\mathbb{R}$ is a constant.

Then $\{ct\}$ is an orthogonal base for $S$ and $\{\frac{\sqrt{3}}{|c|}ct\}$ is an orthonormal base for $S$.

Now, if $q(t)=a_0+a_1t+a_2t^2+a_3t^3$ then one condition I founded to satisfy $\langle ct, q(t)\rangle=0$ is that $c(\frac{a_0}{2}+\frac{a_1}{3}+\frac{a_2}{4}+\frac{a_3}{5})=0$, but I don't know how to write the orthogonal space knowing this, can you help me? Thanks!

Answer 1

You found that orthogonal basis of $S$ is $\{t \sqrt{3}\}$.

And $a_0+a_1t+a_2t^2+a_3t^3 = q(t) \in S^⊥$ if and only if $\frac{a_0}{2}+\frac{a_1}{3}+\frac{a_2}{4}+\frac{a_3}{5}=0$. We can calculate $a_0 = -\frac{2a_1}{3}-\frac{a_2}{2}-\frac{2 a_3}{5}$. Let us find a basis of the null space.

1. Choose $a_1 = 1$, $a_2 = a_3 = 0$ and obtain $a_0 = -2/3$.
2. Choose $a_2 = 1$, $a_1 = a_3 = 0$ and obtain $a_0 = -1/2$.
3. Choose $a_3 = 1$, $a_1 = a_2 = 0$ and obtain $a_0 = -2/5$. So, a basis of the null space is $\{(-2/3, 1, 0, 0), (-1/2, 0, 1, 0), (-2/5, 0, 0, 1)\}$.

This produces a basis of the space $S^⊥$ in the form $\{t - \frac{2}{3}, t^2 - \frac{1}{2}, t^3 - \frac{2}{5}\}$.

Now we can build an orthogonal basis of the space $S^⊥$ using the Gram–Schmidt process.

From this build orthogonal basis we can easily build an orthonormal basis.

P.S. If we run Gram–Schmidt process (and orthonormalization), we obtain the answer: $\left\{3 t-2,6 \sqrt{5} t^2-6 \sqrt{5} t+\sqrt{5}, 20 \sqrt{7} t^3-30 \sqrt{7} t^2+12 \sqrt{7} t-\sqrt{7}\right\}$.

Answer 2

Then,$$S=\{a+bt+ct^{2}+dt^{3}\in P_{3}({\bf R}): a=c=d=0, b\in{\bf R}\},$$ i.e., $S={\rm span}\{t\}$. So a basis for $S$ is given by $\beta_{S}=\{t\}$ and then an orthonormal basis for $S$ is given by $\beta^{or}_{S}=\{t/\|t\|\}$.

Since $S^\perp :=\{q(t)\in P_{3}({\bf R}): \langle p(t),q(t)\rangle =0, p(t)\in S\}$, then

$$S^\perp=\left\{q(t)\in P_{3}({\bf R}):\int_{0}^{1}tq(t)\, {\rm d}t=0 \right\}.$$ Taking $q(t)=a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}\in P_{3}({\bf R})$, then $$\int_{0}^{1}tq(t)\, {\rm d}t=0\implies \frac{a_{0}}{2}+\frac{a_{1}}{3}+\frac{a_{2}}{4}+\frac{a_{3}}{5}=0.$$ So, $$S^\perp=\left\{a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}\in P_{3}({\bf R}): \frac{a_{0}}{2}+\frac{a_{1}}{3}+\frac{a_{2}}{4}+\frac{a_{3}}{5}=0\right\}$$ or logically equivalent $$S^\perp=\{a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}\in P_{3}({\bf R}):30a_{0}+20a_{1}+15a_{2}+12a_{3}=0\}$$ To find a base you could proceed as follows: since $30a_{0}+20a_{1}+15a_{2}+12a_{3}=0$ then in particular $$a_{0}=-\frac{2a_{1}}{3}-\frac{2a_{2}}{4}-\frac{2a_{3}}{5}, \quad a_{1},a_{2},a_{3}\in {\bf R}$$ Then, $$a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}=\left(-\frac{2a_{1}}{3}-\frac{2a_{2}}{4}-\frac{2a_{3}}{5} \right)+a_{1}t+a_{2}t^{2}+a_{3}t^{3}$$ and that can be wrriten as $$a_{1}\left(-\frac{2}{3}+t\right)+a_{2}\left(-\frac{2}{4}+t^{2}\right)+a_{3} \left(-\frac{2}{5}+t^{3}\right). $$ So, a basis for $S^\perp$ is given by $$\beta_{S^\perp}=\left\{ \left(-\frac{2}{3}+t\right),\left(-\frac{2}{4}+t^{2}\right),\left(-\frac{2}{5}+t^{3}\right)\right\}$$ Notice that $\dim S+\dim S^\perp =\dim P_{3}({\bf R})=3+1=4$.