In the book I am studying, the author says:
Since $\phi_q$ is a polynomial of degree $q$, for all $j=1,2, \dots, l$, there exist real numbers $b_{qj}$ such that $$u^j=\sum_{q=0}^{j}b_{qj}\phi_q(u), \forall u\in[-1,1].$$
In this case, $\phi_q$ is actually a Legendre polynomial and the set of $\phi$'s form a basis for the polynomials of degree at most $j$.
I think it is a strong result and I am not able to show it.
How can I argue about the existence of such coefficients $b_{qj}$?
On
Write things in terms of matrices: $$ \begin{bmatrix} \phi_0(u)\\\phi_1(u)\\\phi_2(u)\\\vdots\\\phi_l(u) \end{bmatrix}= \begin{bmatrix} a_{0,0}&0&0&\cdots&0\\ a_{0,1}&a_{1,1}&0&\cdots&0\\ a_{0,2}&a_{1,2}&a_{2,2}&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a_{0,l}&a_{1,l}&a_{2,l}&\cdots&a_{l,l} \end{bmatrix} \begin{bmatrix} u^0\\u^1\\u^2\\\vdots\\u^l \end{bmatrix} $$ Since $\phi_j$ has degree $j$, $a_{j,j}\ne0$. The determinant of a lower triangular matrix is $$ \prod_{j=0}^la_{j,j}\ne0 $$ Thus, we can invert the lower triangular matrix to get a lower triangular matrix: $$ \begin{bmatrix} b_{0,0}&0&0&\cdots&0\\ b_{0,1}&b_{1,1}&0&\cdots&0\\ b_{0,2}&b_{1,2}&b_{2,2}&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ b_{0,l}&b_{1,l}&b_{2,l}&\cdots&b_{l,l} \end{bmatrix} \begin{bmatrix} \phi_0(u)\\\phi_1(u)\\\phi_2(u)\\\vdots\\\phi_l(u) \end{bmatrix} =\begin{bmatrix} u^0\\u^1\\u^2\\\vdots\\u^l \end{bmatrix} $$
What this is saying is that the set $\{\phi_{0}, \phi_{1}, \ldots, \phi_{j}\}$ forms a basis for the vector space of (real) polynomials having degree at most $j$ (for each $0 \leq j \leq l$). Thus the polynomial $x^{j}$ is a linear combination of these polynomials (i.e. the coefficients $b_{qj}$ in question exist).
It is worded a little strangely to say that the equality holds for all $u \in [-1,1]$; really it will hold for all $u \in \mathbb{R}$, since these polynomials are equal everywhere.