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 $\int_0^1 x^k f(x)\mathrm{d} x = 0$ 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?

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 $\int_0^1 x^k f(x)\mathrm{d} x = 0$ for $k=1, 2, \ldots, n$. Prove that $\int_0^1 f^2 = (n+1)^2 (\int_0^1 f)^2$.

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 $\det (H_{n+1} - A) =0$ using the following argument: Suppose that $f(x) = a_0 + a_1 x + \cdots + a_n x^n$. From $\int_0^1 x^k f(x)\mathrm{d} x = 0$ for $k=1, 2, \ldots, n$, we get $n$ equations: $\frac{a_0}{k+1} + \frac{a_1}{k+2}+ \cdots + \frac{a_n}{k+n+1}=0, \, (k = 1, 2, \ldots, n)$, and $\int_0^1 f^2 = (n+1)^2 (\int_0^1 f)^2$ is equivalent to $[1-\frac{1}{(n+1)^2}]a_0 + \frac{a_1}{2} + \cdots + \frac{a_n}{n+1} = 0$. Plus with the $n$ linear equations above, we get a system of homogeneous linear equations which must have a non-zero solution $(a_0, a_1, \ldots, a_n)$. So the coefficient matrix, which is just $H_{n+1} - A$, must be degenerate.

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

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?

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 $\int_0^1 x^k f(x)\mathrm{d} x = 0$ 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?