Matrix given by $a_{ij} = 1/(i+j)$ is non-singular. What is a smart way to see that $$A\stackrel{\cdot}{=} \begin{bmatrix} 1/2 & 1/3 & \cdots & 1/(n+1) \\ 1/3 & 1/4 & \cdots & 1/(n+2) \\ \vdots & \vdots & \ddots & \vdots \\ 1/(n+1) & 1/(n+2) & \cdots & 1/(2n) \end{bmatrix} $$is non-singular?
I computed $\det A$ for $n=1,2$ and $3$ but I failed to see the pattern.
The context is as follows: in $\Bbb R[x] $ with inner product given by $\langle p(x),q(x)\rangle \stackrel{\cdot}{=}\int_0^1p(x)q(x)\,{\rm d}x$, I want to see that if $U\stackrel{\cdot}{=} \{p(x) \in \Bbb R[x]\mid p(0)=0\}$, then $U^{\perp} =\{0\}$ (and hence $U^{\perp\perp}=\Bbb R[x]\neq U$). I took $f(x) \in U^\perp$ and computed  $\langle f(x),x^k\rangle$ for $k\geq 1$. I obtained a homogeneous system for the coefficients of $f(x)$ which has $A $ as the matrix associated. So if $A$ is non-singular I'm done.
 A: Here is one approach:
You know that $x \mapsto x^k$, $k=1,2,...$ are linearly independent and so
the operator $L: \mathbb{R}^n \to \mathbb{R}[x]$ given by $(L \alpha) (x) = \sum_k \alpha_k x^k$ is injective. Hence $L^*L$ is injective, and since
$A= L^*L$ we see that $A$ is invertible. (The inner product specified in
the question is used implicitly throughout this answer.)

Here is an alternative approach to the underlying problem:
Let $s_n(x) = 1-(1-x)^n$. Note that $\|s_n\| \le  1$, and $s_n(x) \to 1_{(0,1]}(x)$ for all $x \in [0,1]$ (and convergence is uniform on
compact subsets of $(0,1]$).
In particular, we have $s_n \cdot p \in U$ for all $p \in \mathbb{R}[x]$, and
$\|p-s_n \cdot p \| \to 0$. It follows that $\overline{U} = \mathbb{R}[x]$.
Since $U^\bot = \overline{U}^\bot$, we see that $U^\bot = \{ 0 \} $.
As a related aside, note that the 'evaluation' operator $Ep = p(0)$ is
not continuous with the given norm.
A: About determinant of $A$, it is a special case of something called Cauchy's double alternant.
In general, given any $2n$ numbers $x_1, \ldots, x_n, y_1, \ldots, y_n$. If one
construct a $n \times n$ matrix with entry $\frac{1}{x_i + y_j}$ at $i^{th}$ row, $j^{th}$ column, one has
$$\det\left[ \frac{1}{x_i + y_j}\right] = \frac{\prod_{1\le i < j\le n}(x_i - x_j)(y_i - y_j)}{\prod_{1\le i, j \le n}(x_i + y_j)}\tag{*1}$$
In particular, this means if $x_i$ and $-y_j$ are $2n$ distinct numbers, the determinant is non-zero and hence the matrix is invertible.
It is not that hard to prove $(*1)$ ourselves, consider the expression
$$\prod_{1\le i, j \le n}(x_i + y_j)\times\det\left[ \frac{1}{x_i + y_j}\right]$$
If one expand the determinant, it is easy to see this expression is a polynomial in $x_i, y_j$. Since the determinant vanishes whenever $x_i$ equals to some $x_j$ or $y_i$ equals to some $y_j$. This polynomial contains $\prod_{1\le i < j\le n}(x_i - x_j)(y_i - y_j)$ as a factor. By matching the degree of polynomial on both sides, we find
$$\det\left[ \frac{1}{x_i + y_j}\right] = \lambda_n \frac{\prod_{1\le i < j\le n}(x_i - x_j)(y_i - y_j)}{\prod_{1\le i, j \le n}(x_i + y_j)}$$
for some constant $\lambda_n$ depends only on $n$.
If one replace $x_1$ by $\epsilon x_1$, $y_1$ by $\epsilon y_1$, send $\epsilon$ to $0$ and look at the limiting behaviors of both sides, one can conclude $\lambda_n = \lambda_{n-1}$. One can verify $\lambda_1 = \lambda_2 = 1$ by hand and hence $\lambda_n = 1$ for all $n$.
Apply this to the Hilbert like matrix $A$ at hand, we get
$$\det\left[\frac{1}{i+j}\right] = \frac{(c_n c_{n+1})^2}{c_{2n+1}} \ne 0
\quad\text{ where }\quad c_n = \prod_{k=1}^{n-1} k!
$$
and hence $A$ is invertible.
