It seems that the matrix you present as an example is $A/2$, not $A$; I assume that $A$ is exactly what is defined (so, e.g., $A_{11}=2$, not $1$).

Well, $A$ is the Gram matrix of the basis $\psi$ with respect to the scalar product $(f,g)=\int_{-1}^1 f(x)g(x)\,dx$; so, if we pass to the orthonormal basis of (normed) Legendre polynomials $\tilde P_n(x)=\sqrt{\frac{2n+1}2}P_n(x)$, say $(\tilde P_0,\dots,\tilde P_{p-1})=\psi S$, then $A=S^{-T}S^{-1}$. Thus $A^{-1}=S^TS$, and hence the polynomial you need is $\sum_{n=0}^{p-1}\tilde P_i^2$.

It seems to be well-known (?) thath the maximum of $|P_n(x)|$ on $[-1,1]$ is attained at the endpoints; thus the claim follows.

It seems that the matrix you present as an example is $A/2$, not $A$; I assume that $A$ is exactly what is defined (so, e.g., $A_{11}=2$, not $1$).

Well, $A$ is the Gram matrix of the basis $\psi$ with respect to the scalar product $(f,g)=\int_{-1}^1 f(x)g(x)\,dx$; so, if we pass to the orthonormal basis of (normed) Legendre polynomials $\tilde P_n(x)=\sqrt{\frac{2n+1}2}P_n(x)$, say $(\tilde P_0,\dots,\tilde P_{p-1})=\psi S$, then $A=S^{-T}S^{-1}$. Thus $A^{-1}=SS^T$, and hence the polynomial you need is $P(x)=(\psi S)(\psi S)^T=\sum_{n=0}^{p-1}\tilde P_i^2(x)$.

It seems to be well-known (?) thath the maximum of $|P_n(x)|$ on $[-1,1]$ is attained at the endpoints; thus the claim follows.

It seems that the matrix you show is $A/2$; I assume that $A$ is exactly what is defined (so, e.g., $A_{11}=2$, not $1$).

Well, $A$ is the Gram matrix of the basis $\psi$ with respect to the scalar product $(f,g)=\int_{-1}^1 f(x)g(x)\,dx$; so, if we pass to the orthonormal basis of (normed) Legendre polynomials $\tilde P_n(x)=\sqrt{\frac{2n+1}2}P_n(x)$, say $(\tilde P_0,\dots,\tilde P_{p-1})=\psi S$, then $A=S^{-T}S^{-1}$. Thus $A^{-1}=S^TS$, and hence the polynomial you need is $\sum_{n=0}^{p-1}\tilde P_i^2$.

It seems to be well-known (?) thath the maximum of $P_n$ is attained at the endpoints of the segment $[-1,1]$; thus the claim follows.

It seems that the matrix you present as an example is $A/2$, not $A$; I assume that $A$ is exactly what is defined (so, e.g., $A_{11}=2$, not $1$).

Well, $A$ is the Gram matrix of the basis $\psi$ with respect to the scalar product $(f,g)=\int_{-1}^1 f(x)g(x)\,dx$; so, if we pass to the orthonormal basis of (normed) Legendre polynomials $\tilde P_n(x)=\sqrt{\frac{2n+1}2}P_n(x)$, say $(\tilde P_0,\dots,\tilde P_{p-1})=\psi S$, then $A=S^{-T}S^{-1}$. Thus $A^{-1}=S^TS$, and hence the polynomial you need is $\sum_{n=0}^{p-1}\tilde P_i^2$.

It seems to be well-known (?) thath the maximum of $|P_n(x)|$ on $[-1,1]$ is attained at the endpoints; thus the claim follows.