Finding the determinant of a tridiagonal matrix

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

• A little experimenting suggests that the answer is $(x-n)^{n+1}.$ – saulspatz Apr 22 at 13:19
• 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. – darij grinberg Apr 22 at 20:32
• Something is telling me that the nilpotency must come from the $\mathfrak{sl}_2$-module of degree-$n$ homogeneous binary forms. – 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 $$\det A=(x-n)^{n+1}$$.
• 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 $P$! – darij grinberg Apr 23 at 15:50
• 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). – darij grinberg Apr 23 at 16:05