On the determinant of a Toeplitz-Hessenberg matrix

Question

I am having trouble proving that

$$\det \begin{pmatrix} \dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\ \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\ \dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\ \dfrac{1}{n!} & \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} \end{pmatrix} =\dfrac{1}{n!}. $$

Answer 1

Hint. In general, let $d_0=d_1=1$ and let $(a_k)_{k=1,2,\ldots}$ be any sequence of numbers. For every $n\ge2$, denote by $d_n$ the determinant of the $n\times n$ Toeplitz-Hessenberg matrix $$ \begin{pmatrix} a_1 &1 &0 &0 &\cdots &0\\ a_2 &a_1 &1 &0 &\cdots &0\\ a_3 &a_2 &a_1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\ a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1 &1\\ a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1 \end{pmatrix}. $$ If one expands the determinant by the first column, one obtains $$ d_n=-\sum_{k=1}^n(-1)^ka_kd_{n-k}. $$

Answer 2

Hints

Prove it by induction.

At each step, expand by minors along the top row.

At the end, think about the binomial theorem.

Answer 3

HINT.-By property of determinants, we lower the order of n to (n-1) as follows $$\Delta_n=\det\begin{pmatrix} 1 &1 &0 &0 &\cdots &0\\ a_2 &1 &1 &0 &\cdots &0\\ a_3 &a_2 &1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\ a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1 &1\\ a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1 \end{pmatrix}$$

$$\Delta_n=\det\begin{pmatrix} 1 &0 &0 &0 &\cdots &0\\ a_2 &1-a_2 &1 &0 &\cdots &0\\ a_3 &a_2-a_3 &1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\ a_{n-1} &a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\ a_n &a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1 \end{pmatrix}$$ $$\Delta_n=\det\begin{pmatrix}1-a_2 &1 &0 &\cdots &0\\ a_2-a_3 &1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\vdots &\\ a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\ a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1 \end{pmatrix}$$

On the other hand one has $\Delta_n=\dfrac{1}{n!}=\dfrac{1}{n}\dfrac{1}{(n-1)!}=\dfrac 1n\Delta_{n-1}$ so we have to prove that

$$\det\begin{pmatrix}1-a_2 &1 &0 &\cdots &0\\ a_2-a_3 &1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\vdots &\\ a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\ a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1 \end{pmatrix}=\dfrac 1n\Delta_{n-1}$$ $$\dfrac 1n\Delta_{n-1}=\det\begin{pmatrix}\dfrac 1n &1 &0 &\cdots &0\\ \dfrac{1}{n}a_2 &1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\vdots &\\ \dfrac 1na_{n-2} &a_{n-3} &\cdots &1 &1\\ \dfrac 1na_{n-1} &a_{n-2} &a_{n-3} &\cdots &1 \end{pmatrix}$$

Note that the columns in these two last determinants are all equal excepting the first.

Can you apply comfortably induction now to prove that both two last determinants are equal?