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Can someone give me a hint how to solve \begin{align*} |A|=\begin{vmatrix} x & 1 & 0 & 0 & \cdots & 0 & 0 \\ n-1 & x & 2 & 0 & \cdots & 0 & 0 \\ 0 & n-2 & x & 3 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \\ \end{vmatrix}? \end{align*}

By adding to row 1 the rows from 2 to n ,I can see that $|A|$ has $x+n-1$ as a factor. And by trying $n=2,3,4$, I can deduce that $$|A|=(x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-1^2)$$ if $n$ is even, and $$|A|=(x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-2^2)x$$ if $n$ is odd. I tried hard to give a proof, but without any progress.

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  • $\begingroup$ Note that it would suffice to show that the determinant of $A$ is $0$ when $x = \pm (n-1), \pm (n-3), \pm (n-5),\dots$. I'm not sure how you would go about showing that though. $\endgroup$ – kccu Apr 21 at 16:32
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    $\begingroup$ This is the characteristic polynomial of a Kac matrix. You may see this question for more details. $\endgroup$ – user1551 Apr 29 at 14:36
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Denote $|A|=D_n(x)$. By adding the rows numbered $2, 3, \ldots, n$ to row $1$, we get \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & x+n-1 & x+n-1 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ n-1 & x & 2 & 0 & \cdots & 0 & 0 \\ 0 & n-2 & x & 3 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{vmatrix} \end{align*} Subsequently we add the rows numbered $3, 4, \ldots, n$ to row $2$, and obtain \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & x+n-1 & x+n-1 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ n-1 & x+n-2 & x+n-1 & x+n-1 & \cdots &x+n-1 & x+n-1 \\ 0 & n-2 & x & 3 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{vmatrix} \end{align*} $\cdots$ and so on, by adding the rows numbered $k+1, k+2, \ldots, n$ to row $k$ ($k=1,2,\cdots,n-1$) successively, we finally get \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & x+n-1 & x+n-1 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ n-1 & x+n-2 & x+n-1 & x+n-1 & \cdots &x+n-1 & x+n-1 \\ 0 & n-2 & x+n-3 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ 0 & 0 & n-3 & x+n-4 & \cdots & x+n-1 & x+n-1 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & x+1 & x+n-1 \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{vmatrix} \end{align*} In this determinant the elements above the diagonal are all $x+n-1$.We successively subtract the $k$-th column from the $k+1$-th column, $k=n-1,n-2,\cdots,1$, and finally get \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ n-1 & x-1 & 1 & 0 & \cdots &0 & 0 \\ 0 & n-2 & x-1 & 2 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x-1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & x-1 & n-2 \\ 0 & 0 & 0 & 0 & \cdots & 1 & x-1 \end{vmatrix} \end{align*} By expanding the above determinant along its first row, we obtain \begin{align*} D_n(x)=&(x+n-1)\begin{vmatrix} x-1 & 1 & 0 & \cdots &0 & 0 \\ n-2 & x-1 & 2 & \cdots & 0 & 0 \\ 0 & n-3 & x-1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x-1 & n-2 \\ 0 & 0 & 0 & \cdots & 1 & x-1 \end{vmatrix}\\ =&(x+n-1)D_{n-1}(x-1) \end{align*} Using the above recurrence relation, we can get \begin{align*} D_n(x)=&(x+n-1)(x+n-3)D_{n-2}(x-2)\\ =&(x+n-1)(x+n-3)\cdots (x+n+1-2k)D_{n-k}(x-k)\\ =&(x+n-1)(x+n-3)\cdots (x-n+3)D_{1}(x-n+1)\\ =&(x+n-1)(x+n-3)\cdots (x-n+3)(x-n+1)\\ =&\prod_{k=1}^{n}\big(x+n+1-2k\big)\\ =&\begin{cases} (x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-1^2),~~~\text{when $n$ is even}\\ (x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-2^2)x,~\text{when $n$ is odd.} \end{cases} \end{align*}

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  • $\begingroup$ Very nice induction proof! $\endgroup$ – darij grinberg Apr 29 at 20:27
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We set $x=0$ and show that $A$ has eigenvalues $\pm(n-1),\pm(n-3),\ldots$ by finding a matrix that diagonalizes it. The solution of the problem follows easily from there.

Let $P$ be the Pascal matrix defined by $$p_{ij} = {{j-1}\choose{i-1}} = \frac{(j-1)^{\underline{i-1}}}{(i-1)!}\,,$$ where we have used generalization of binomial coefficient, defined using falling factorials. For example, when $n=4$ we have $$P=\begin{bmatrix}1&1&1&1\\ 0&1&2&3\\ 0&0&1&2\\ 0&0&0&1\end{bmatrix}\,.$$ Let $S$ be diagonal matrix defined by $$s_{ii} = 2^{n-i}{n-1\choose i-1}\,.$$ Then matrix $$V = P^{-1}SP^T$$ is such that $V^{-1}AV$ is diagonal, with eigenvalues sorted in descending order.

Proof. To prove this claim, let $L$ be the lower triangle and $U$ be the upper triangle of matrix $A$. Also, let $D$ be diagonal matrix defined by $d_{ii}=n+1-2i$, so that its diagonal elements are eigenvalues sorted in descending order.

We start with equivalence $$V^{-1}AV=D \;\Leftrightarrow\; S^{-1}PAP^{-1}S = P^TDP^{-T}\,.$$

To simplify the last equality we will use the following three equalities $$PUP^{-1}=U\,,\quad PLP^{-1}=D+L-U\,,\quad PDP^{-1}=D-2U\,.$$ One can prove them by multiplying them from the right by $P$ and by comparing the resulting left and right hand side, using properties of binomial coefficient.

We sum the first two equalities to obtain $$PAP^{-1}=D+L\,.$$ We use the last and the first equality to obtain $$D=P^{-1}DP-2P^{-1}UP = P^{-1}DP-2U\,.$$ Adding $2U$ to the both sides of this equality and transposing gives $$P^TDP^{-T}=D+2U^T\,.$$

Now, we have $$V^{-1}AV=D \;\Leftrightarrow\; S^{-1}(D+L)S=D+2U^T \;\Leftrightarrow\; S^{-1}LS=2U^T \,.$$ The last equation can be seen to hold by multiplying it from the left by $S$ and comparing the resulting left and right hand side. This ends the proof.

The idea for solution came from the answer to this question.

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  • $\begingroup$ Thanks for your reply,it's not easy to think of constructing such a matrix $P$,I will read it carefully.By the way,below I have posted a solution to this question,which needs some tricks of determinant. $\endgroup$ – mbfkk Apr 29 at 14:33

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