How to show a Hilbert matrix is invertible? I got the matrix for the standard inner product space on polynomial space $\mathbb{P}_n$  as
$$H_n=\begin{bmatrix}1&1/2&1/3&\cdots&1/(n+1)\\1/2&1/3&1/4&\cdots&1/(n+2)\\\cdots&\cdots&\cdots&\cdots&\cdots\\1/(n+1)&1/(n+2)&1/(n+3)&\cdots&1/(2n+1)\end{bmatrix}.$$
We know matrix representation of inner product is invertible. But I want to show explicitly that the obtained matrix is invertible. How to show that?
 A: Define $\langle.,,\rangle$ on $\mathbb{P}_n$ as
$$
\langle p(x), q(x)\rangle = \int_0^1 p(x)q(x)dx.
$$
It is easy to check that $\langle.,,\rangle$ is an inner product.
Notice that
$$
H_n=\begin{bmatrix}
\langle 1,1\rangle&\langle 1,x\rangle&\langle 1,x^2\rangle&\cdots&\langle 1,x^n\rangle\\
\langle x,1\rangle&\langle x,x\rangle&\langle x,x^2\rangle&\cdots&\langle x,x^n\rangle\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
\langle x^n,1\rangle&\langle x^n,x\rangle&\langle x^n,x^2\rangle&\cdots&\langle x^n,x^n\rangle\\
\end{bmatrix}$$
is the Gram matrix of $1,x,x^2,\dots,x^n$ with $\langle.,,\rangle$. Now, determinant of a Gram matrix is non-zero if and only if the vectors whose inner products are used in its construction are linearly independent. However, $1,x,x^2,\dots,x^n$ is a basis of $\mathbb{P}_n$ and hence linearly independent. Therefore, $\det H_n\ne 0$ and so we conclude that $H_n$ is invertible.
A: It suffices to show that the equation $H_nx = 0$ has the unique solution $x = 0$ (i.e. that its columns are linearly independent). So, suppose that $x = (x_1,\dots,x_{n}, x_{n+1})$ is such a solution. It follows that $x^T(H_n x) = 0$, which is to say that
\begin{align}
0 & = x^TH_n x = \sum_{i,j = 1}^{n+1} H_n[i,j] x_i x_j 
\\ & = \sum_{i,j = 1}^{n+1}  x_i x_j \int_0^1 t^{i-1} t^{j-1}\, dt
\\ & = \int_0^1 \sum_{i,j = 1}^{n+1} x_ix_j \,t^{i-1}t^{j-1}\,dt 
= \int_0^1 (x_1 + x_2 t + \cdots + x_{n+1} t^{n})^2\,dt.
\end{align}
This integral can only be zero if $x_1 + x_2 t + \cdots + x_n t^{n-1}$ is the zero-function over $[0,1]$ (in general, the integral of a continuous non-negative function over an interval is zero iff that function is identically zero). However, this only occurs if $x_1 = \cdots = x_n = 0$, which is to say that $x = 0$, which is what we wanted to show.
