I was trying to solve this exercise but I can't get to answer given in book:
Find the singular values of the operator D over $R_2[x]$ (That is polynomials with degree equal or less than 2) define as D(p)=p' where $p \in R_2[x]$ and p' is the derivate of the polynomial. And the inner product over $R_2[x]$ is :
$\langle p,q \rangle = \displaystyle{\int_{-1}^{1}p(x)q(x)dx}$.
So this is the way I'd tried to solve it:
First of all I took the base {1,x,$x^2$} and used gram-schmidt process to orthonormalised the set of vectors, {$\sqrt{\frac{1}{2}},\sqrt{\frac{3}{2}}x,\sqrt{\frac{45}{8}}(x^2-\frac{1}{3})$}, and then I found the matrix of D, given that $D(\sqrt{\frac{1}{2}})=0$, $D(\sqrt{\frac{3}{2}}x)=\sqrt{\frac{3}{2}}$ and $D(\sqrt{\frac{45}{8}}(x^2-\frac{1}{3}))=2\sqrt{\frac{45}{8}}x=3\sqrt{\frac{5}{2}}x$:
$\mathcal{M}(D)=\begin{pmatrix} 0 & \sqrt{\frac{3}{2}} & 0\\ 0 & 0 & 3\sqrt{\frac{5}{2}}\\ 0 & 0 & 0 \end{pmatrix}$
Now the matrix of the adjoint operator is the transpose matrix of $\mathcal{M}(D)$, that is:
$\mathcal{M}(D^*)=\mathcal{M}(D)^{T}=\begin{pmatrix} 0 & 0 & 0\\ \sqrt{\frac{3}{2}} & 0&0\\ 0 & 3\sqrt{\frac{5}{2}} & 0 \end{pmatrix}$. So:
$\mathcal{M}(DD^*)=\mathcal{M}(D)\mathcal{M}(D^*)=\begin{pmatrix} \frac{3}{2} & 0 & 0\\ 0 & \frac{45}{2} &0\\ 0 & 0 & 0 \end{pmatrix}$.
So the singular values of D would be the positive squre of the eigenvalues of the matrix $\mathcal{M}(DD^*)$, that is the singular values of D would be $\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{45}{2}}$. But according to the book the answer is wrong.
There might be just a silly mistake doing the maths but I can't see where is it. I would really appreciate any advice or hint you could give me.
Thanks