Find the determinant whose result is $(x-n)^{n+1}$

Question

Find the determinant $$ \left|\begin{array}{cccccc}{x} & {1} & {} & {} & {} & {} \\ {-n} & {x-2} & {2} & {} & {} & {} \\ {} & {-(n-1)} & {x-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-2} & {x-2 n+2} & {n} \\ {} & {} & {} & {} & {-1} & {x-2 n}\end{array}\right| $$

I know the answer is $(x-n)^{n+1}$, and I tried to find it with Gaussian elimination but failed. (That is to reduce the matrix to a matrix with all $(x-n)$ on the diagonal line.)

How to find the determinant? Any method will be appreciated.

Answer 1

Here is a different approach using Jordan form.

Setting $x=-\lambda+n$ in the given determinant, the given issue is equivalent to the following one : establish that the characteristic polynomial of:

$$M=\left(\begin{array}{cccccc}{n} & {1} & {} & {} & {} & {} \\ {-n} & {n-2} & {2} & {} & {} & {} \\ {} & {-n+1} & {n-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-2} & {-n+2} & {n} \\ {} & {} & {} & {} & {-1} & {-n}\end{array}\right)$$

is

$$(-\lambda)^{n+1}.$$

This will be obtained if we establish that the Jordan decomposition

$$M=VJV^{-1}\tag{1}$$

gives a matrix $J$ with $J_{ij}=\delta(i-j+1)$ (the only non-zero entries of $J$ are "ones" on the first upper diagonal).

Let us show it, for the sake of simplicity, on the case $n=3$ (the general case being in direct line with this particular case).

$$M=\left(\begin{array}{rrrr} 3 & 1 & 0 & 0\\ -3 & 1 & 2 & 0\\ 0 & -2 & -1 & 3\\ 0 & 0 & -1 & -3 \end{array}\right), \ J=\left(\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{array}\right), \ V=\left(\begin{array}{rrrr} 6 & 6 & 3 & 1\\ -18 & -12 & -3 & 0\\ 18 & 6 & 0 & 0\\ -6 & 0 & 0 & 0 \end{array}\right)$$

Proof : $M$ and $J$ being similar due to relationship (1), they share the same characteristic polynomial which is clearly $(-\lambda)^n.$

Remarks :

1) The general case is not difficult by itself, but is long (and rather uninteresting) to write down. In the general case, as in the particular case, the columns of $V$, starting from the right, are proportional to the successive rows of Pascal's triangle (with alternate signs) .

2) Matrix $M$ belongs to a category called skew-centronormal matrices with a certain litterature around them.

3) Matrix $M$ is very ill-conditionned.

Answer 2

Here's a possible method:

From up to down, add the row above to each row:

$$ D_n=\left|\begin{array}{cccccc}{x} & {1} & {} & {} & {} & {} \\ {x-n} & {x-1} & {2} & {} & {} & {} \\ {x-n} & {x-n} & {x-2} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {x-n} & {x-(n-1)} & {n} \\ {x-n} & {x-n} & {} & {} & {x-n} & {x- n}\end{array}\right| $$ From left to right, subtract each column from the right column, thus $$ D_n=\left|\begin{array}{cccccc}{x-1} & {1} & {} & {} & {} & {} \\ {-(n-1)} & {x-1-2} & {2} & {} & {} & {} \\ {} & {-(n-2)} & {x-1-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-1} & {x-1-2(n-1)} & { } \\ {} & {} & {} & {} & { } & {x- n}\end{array}\right| =(x-n)D_{n-1}$$ As $D_0=x-n$, we deduce that $D_n=(x-n)^{n+1}$.