Let $H_{n+1}$ denote the $(n+1)\times (n+1)$ Hilbert matrix, i.e., the $(i, j)$-entry of $H_{n+1}$ is $(i+j-1)^{-1}$ with $1 \leq i, j \leq n+1$. Let $A$ denote an $(n+1)\times (n+1)$ matrix whose entries are all zero but the upper-left corner element is $(n+1)^{-2}$. Then $H_{n+1} - A$ has determinant zero. 

I obtained this result from an exercise in my Analysis textbook. That exercise is as follow. Let $f(x)$ be a polynomial of degree $n$, such that $\int_0^1 x^k \mathrm{d} x = 0$ for $k=1, 2, \ldots, n$. Prove that $\int_0^1 f^2 = (n+1)^2 (\int_0^1 f)^2$. 

This exercise is not hard. But more importantly, such a polynomial do exist. (One can find a polynomial $F(x)$ of degree $2n$ such that $F^{(i)}(0) = F^{(i)}(1)=0$ for $0 \leq i \leq n-1$ and $f(x) = F^{(n)}(x)$.) And hence I derive that $\det (H_{n+1} - A) =0$. 

My question is, can we prove $\det (H_{n+1} - A) =0$ using other method? For instance, is there any linear algebra trick can guarantee this?