$$\begin{vmatrix}x&1&0&0&⋯\\-n&x-2&2&0&⋯\\0&-(n-1)&x-4&3&⋯\\⋮&⋱&⋱&⋱&⋮\\0&⋯&-2&x-2(n-1)&n\\0&0&⋯&-1&x-2n\end{vmatrix}_{(n+1)×(n+1)}$$Find the value of the above determinant.

This problem comes from an advanced algebra book. I want to solve it with elementary transformation knowledge. I have been trying to solve it for a long time.

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    $\begingroup$ A little experimenting suggests that the answer is $(x-n)^{n+1}.$ $\endgroup$ – saulspatz Apr 22 at 13:19
  • $\begingroup$ Note that @saulspatz's observation means (by considering eigenvalues) that if you set $x = n$ in your matrix, then the resulting matrix is nilpotent. This may be a simpler thing to prove. $\endgroup$ – darij grinberg Apr 22 at 20:32
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    $\begingroup$ Something is telling me that the nilpotency must come from the $\mathfrak{sl}_2$-module of degree-$n$ homogeneous binary forms. $\endgroup$ – darij grinberg Apr 22 at 20:43

Edit. Call your matrix $A$ and let $L$ be the zero-indexed Pascal matrix defined by $$ l_{ij}=\begin{cases} \binom{n-j}{i}&\text{ when }\ 0\le i\le n-j\le n,\\ 0&\text{ otherwise}. \end{cases} $$ It is known that $(L^{-1})_{ij}=(-1)^{i+j}l_{ij}$. E.g. when $n=5$, $$ L=\pmatrix{1&0&0&0&0&0\\ 5&1&0&0&0&0\\ 10&4&1&0&0&0\\ 10&6&3&1&0&0\\ 5&4&3&2&1&0\\ 1&1&1&1&1&1}, \ L^{-1}=\pmatrix{1&0&0&0&0&0\\ -5&1&0&0&0&0\\ 10&-4&1&0&0&0\\ -10&6&-3&1&0&0\\ 5&-4&3&-2&1&0\\ -1&1&-1&1&-1&1}. $$ One may verify that $A-(x-n)I_{n+1}=L^{-1}NL$ for some nilpotent matrix $N$. More specifically, one may verify that $L(A-(x-n)I_{n+1})=NL$ where $$ N=\pmatrix{0&1\\ &0&2\\ &&\ddots&\ddots\\ &&&\ddots&n\\ &&&&0}. $$

In other words, if $V$ is the vector space of polynomials in $y$ of degrees $\le n$ and $D,g,g^{-1}:V\to V$ are the linear operators \begin{aligned} D(p)(y)&=p'(y),\\ g(p)(y)&=(1+y)^np\left(\frac{y}{1+y}\right),\\ g^{-1}(p)(y)&=(1-y)^np\left(\frac{y}{1-y}\right), \end{aligned} then $A-(x-n)I_{n+1}$ is the matrix representation of the linear map $f=g^{-1}\circ D\circ g$ with respect to the ordered basis $(1,y,y^2,\ldots,y^n)$. (More explicitly, $f(p)(y)=n(1-y)p(y)+(1-y)^2p'(y)$, but this formula is unimportant here.) Since $D^n=0$, $f^n=g^{-1}\circ D^n\circ g$ is also zero. Hence $f$ and $A-(x-n)I_{n+1}$ are nilpotent and in turn $\det A=(x-n)^{n+1}$.

  • $\begingroup$ I have checked your $P(A-(x-n)I_{n+1})=JP$ identity, at least away from the borders of the matrix. It boils down to $\left(n-j-i+1\right)\left(n-2j\right) + j\left(n-j+1\right) - \left(n-j-i\right)\left(n-j-i+1\right) = i\left(n-i+1\right)$ for all $i$ and $j$, which is painful but not exactly difficult to verify. Good job identifying $P$! $\endgroup$ – darij grinberg Apr 23 at 15:50
  • $\begingroup$ Now I am convinced that the matrix $A - \left(x-n\right) I_{n+1}$ represents the "finite difference" operator $f\left(t\right) \mapsto f\left(t\right) - f\left(t-1\right)$ on the degree-$\leq n$ subspace of the polynomial ring $\mathbb{Q}\left[t\right]$ with respect to some basis. The question is just what the basis is. The basis should likely consist of $n+1$ degree-$n$ polynomials (it cannot be triangular for the degree, as otherwise the operator too would be triangular). $\endgroup$ – darij grinberg Apr 23 at 16:05

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