Find the singular values of $T: p(x)\mapsto xp'(x)+2x^2p''(x)$ I would appreciate help with this problem I am self-studying this problem:


Find the singular values of the operator $T\in \mathcal P_2(\mathbb{C}): p(x)\mapsto xp'(x)+2x^2p''(x)$. The inner-product on $\mathcal P_2(\mathbb{C})$ is defined as $\langle p,q \rangle:=\int_{-1}^{1} p(x)\overline{q(x)}dx$.


I know the singular values are the eigenvalues of $\sqrt{T^*T}$, where $T=S \sqrt{T^*T}$ and $S$ is an isometry.
Thanks
EDIT I am a bit skeptical of the hint to construct a matrix, for one, because I don't see how that will utilize the defined inner-product. I think for a start, I should write:
$\langle Tp,q\rangle= \int_{-1}^{1} Tp(x)\overline{q(x)}dx= \int_{-1}^{1} p(x)\overline{T^*q(x)}dx=\langle p,T^*q\rangle$, where $Tp(x)=xp'(x)+2x^2p''(x)$
But, assuming this is on the right tract, I don't know how to proceed. Thanks again
 A: Start with the basis $\{1, x, x^2\}$ for $\mathcal{P}_2(\mathbb{C})$. Performing Gram-Schmidt yields the following orthonormal basis for $\mathcal{P}_2(\mathbb{C})$: $$\mathcal{B} = \left\{ \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}x, \sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right\}_.$$
Let $A$ denote the matrix of $T$ with respect to $\mathcal{B}$. We then compute that:
$$A = \begin{pmatrix}
0 & 0 & \sqrt{45}\\
0 & 1 & 0\\
0 & 0 & 6
\end{pmatrix} = \begin{pmatrix}
0 & 0 & 3\sqrt{5}\\
0 & 1 & 0\\
0 & 0 & 6
\end{pmatrix}$$
Since $\mathcal{B}$ is an orthonormal basis, the matrix of $T^*$ with respect to $\mathcal{B}$ is the conjugate transpose of $A$ (Relevant link if you feel shaky about this). Therefore the matrix of $T^*T$ with respect to $\mathcal{B}$ is:
$$\overline{A^T}A = \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
3\sqrt{5} & 0 & 6
\end{pmatrix}\begin{pmatrix}
0 & 0 & 3\sqrt{5}\\
0 & 1 & 0\\
0 & 0 & 6
\end{pmatrix} = \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 81 \\
\end{pmatrix}_.$$
Finally, the matrix of $\sqrt{T^*T}$ with respect to $\mathcal{B}$ is then:
$$\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 9
\end{pmatrix}_.$$
Thus the singular values of $T$ are $0,1, \text{ and } 9$.


Per request: How to find the matrix $A$.
To do so, we find the image of each element of $\mathcal{B}$ under $T$, then express the image as a linear combination of the elements of $\mathcal{B}$. The coefficients of this linear combination become entries of the matrix.

$$T\left(\frac{1}{\sqrt{2}}\right) = 0 = 0\left(\frac{1}{\sqrt{2}}\right) + 0 \left(\sqrt{\frac{3}{2}}x \right) + 0 \left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right)_.$$
This corresponds to the first column of $A$ being $\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}_.$
$$T\left(\sqrt{\frac{3}{2}}x\right) = x\sqrt{\frac{3}{2}} + 2x^2(0) = 0 \left(\frac{1}{\sqrt{2}}\right) + 1 \left(\sqrt{\frac{3}{2}}x\right) + 0 \left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right)_.$$
This corresponds to the second column of $A$ being $\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}_.$
\begin{align*}
T\left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right) &= x\left(2\sqrt{\frac{45}{8}}x\right) + 2x^2\left(2\sqrt{\frac{45}{8}}\right)\\
&= 6\sqrt{\frac{45}{8}}x^2 \\
&= \sqrt{45}\left(\frac{1}{\sqrt{2}}\right) + 0 \left(\sqrt{\frac{3}{2}}x\right) + 6\left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right)_.
\end{align*}
This corresponds to the third column of $A$ being $\begin{pmatrix} \sqrt{45} \\ 0 \\ 6 \end{pmatrix}_.$
