How to do SVD of the diﬀerentiation operator of $P_{2}$ Find the singular values of the differentiation operator $D \colon P_{2}  \rightarrow P_{2}$ where $P_{2}$ is the vector space of real polynomials of degree less than or equal to $2$, and the inner product on $P_{2}$ is given by 
$$ \left<f,g \right> :=\int_{-1}^{1}f(t)g(t) \,dt. $$
Using the brute-force method, I cannot even find the adjoint of this map.
I have just came up with the idea that the only eigenvalue of the differential operator is $0$, as the conjugation of $0$, $0$ is the only eigenvalue of the $D^*$. But I cannot say anything about the composition $D^{*}D$...
So could someone provide some ideas of solving this? Thanks so much.
 A: You can start with finding an orthonormal basis for $P_2$ by applying Gram-Schmidt to the standard basis $(1,x,x^2)$. Using the fact that the integral of an odd function on $[-1,1]$ vanishes, we get:
$$ e_1 = \frac{1}{\|1\|} = \frac{1}{\sqrt{2}}, \\
e_2 = \frac{x - \left< x, e_1 \right> e_1}{x - \left< x, e_1 \right> e_1} = \frac{x}{\|x\|} = \sqrt{\frac{3}{2}} x, \\
e_3 = \frac{x^2 - \left< x^2, e_2 \right> e_2 - \left< x^2, e_1 \right> e_1}{\|x^2 - \left< x^2, e_2 \right> e_2 - \left< x^2, e_1 \right> e_1\|} = \frac{x^2 - \frac{1}{2} \int_{-1}^1 x^2 \, dx}{\| x^2 - \frac{1}{2} \int_{-1}^1 x^2 \, dx \|} = \frac{x^2 - \frac{1}{3}}{\| x^2 - \frac{1}{3} \|} = \sqrt{\frac{45}{8}} \left( x^2 - \frac{1}{3} \right).$$
Then, we calculate:
$$
D(e_1) = 0,\\
D(e_2) = \sqrt{\frac{3}{2}} = \sqrt{3} e_1, \\
D(e_3) = \sqrt{\frac{45}{2}}x = \sqrt{15} e_2.
$$
From here, since the $e_i$ are orthonormal, we already see that the singular values of $D$ are $\sqrt{15}, \sqrt{3}, 0$. Alternatively, we can see that with respect to $\mathcal{B} = (e_1, e_2, e_3)$, the operator $D$ is represented by the matrix
$$ [D]_{\mathcal{B}} = \begin{pmatrix} 0 & \sqrt{3} & 0 \\ 
0 & 0 & \sqrt{15} \\
0 & 0 & 0 \end{pmatrix}. $$
and $D^{*}$ is represented by the transpose
$$ [D^{*}]_{\mathcal{B}} = [D]_{\mathcal{B}}^t = \begin{pmatrix} 0 & 0 & 0 \\ 
\sqrt{3} & 0 & 0 \\
0 & \sqrt{15} & 0 \end{pmatrix}. $$
Hence,
$$ [D^{*}D]_{\mathcal{B}} = [D^{*}]_{\mathcal{B}} [D]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 15 \end{pmatrix} $$
which also shows that the eigenvaleus of $\sqrt{D^{*}D}$ which are the singular values of $D$ are $\sqrt{15}, \sqrt{3}, 0$.
