Transformation matrix of the derivative given the Legendre-basis for polynomials

Question

In the space of polynomials with degree $n\leq 3$ $V$ equipped with the inner product induced by $$\langle f,g\rangle:=\int_{-1}^{1}f(t)g(t)\mathrm{d}t$$ the accordingly normed Legendre polynomials $$\left\{\frac{1}{\sqrt{2}},\frac{\sqrt{6}}{2}x,\frac{\sqrt{10}}{4}(3x^2-1), \frac{\sqrt{14}}{4}(5x^3-3x)\right\}$$ form an orthonormal basis. As an exercise, I had to give the adjoint of the derivative $p\mapsto p'.$

So, I am currently trying to first get the Transformation matrix of the derivative according to the given basis, and then use transposition to find the adjoint. However I have problems in expressing the image of the last listed basis vector, $\frac{14\sqrt{14}}{4}x^2-\frac{3\sqrt{14}}{4}$, in terms of only the third, $\frac{\sqrt{10}}{4}(3x^2-1)$, as they are not linearly dependent. I therefore think I made an error, as I can not get the typical diagonal in the transformation matrix, but i cannot find it.

Thank you for any help in advance.

Answer 1

In your calculations, it seems that all your diagonal entries should be $0$. That is, letting $\{e_0, e_1, e_2, e_3\}$ be the onb you gave above, you will find that the $i$th coefficient of $\frac{d}{dx} e_i$ expanded in that onb is zero.

For example:

$$ \frac{d}{dx} \left\{ \frac{\sqrt{14}}{4}(5x^3 - 3x) \right\} = \frac{5\sqrt{14}}{4}\left(3x^2 - \frac{3}{5}\right) = \frac{5\sqrt{14}}{4}\left(3x^2 - 1\right) + \frac{5\sqrt{14}}{4}\cdot \frac{2}{5} \\= a\, e_0 + b\, e_1 + c \, e_2 + k\, e_3 $$ has $a = - \frac{\sqrt{14}/2}{1/\sqrt{2}}$, $b = 0$, $c = \frac{5\sqrt{14}/4}{\sqrt{10}/4}$ and $k = 0$.

Answer 2

A straightforward way to proceed is to apply a change of-basis to the matrix of the derivative operator relative to the standard basis, i.e., to the matrix $$D=\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}.$$ Letting $L$ be the matrix with the coefficients of the Legendre polynomials as its columns, we have $D'=L^{-1}DL$ relative to your basis. If you want to avoid direct matrix inversion and multiplication, you can compute this by row-reducing the augmented matrix $M=\left[\begin{array}{c|c}L&DL\end{array}\right]$, which will produce $\left[\begin{array}{c|c}I&L^{-1}DL\end{array}\right]$. The right side of $M$ consists of the coefficients of the derivatives of the Legendre polynomials, so you can avoid a matrix multiplication there, too, if you’re so inclined.