# Determinant of a polynomial: Exercise from chapter 1, "Linear Algebra" by Georgi E. Shilov

I am solving the exercise from "Linear Algebra" by Georgi E. Shilov. The question asks to prove the following:

\begin{align*} \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots &\vdots \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2}\\ x_1^{n} & x_2^{n} & \cdots & x_n^{n}\\ \end{vmatrix} = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots &\vdots \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2}\\ x_1^{n-1} & x_2^{n-2} & \cdots & x_n^{n-1}\\ \end{vmatrix} \times \sum_{k=1}^{n}{x_k} \end{align*}

Initially I thought this was a Vandermonde Determinant, but the power of the second last row is $$(n - 2)$$ after which we skip $$(n - 1)$$ and go to $$n$$.

Second, I don't understand the pattern of the last row of the RHS. Is it alternating $$(n - 1)$$'s and $$(n - 2)$$'s?

Hence, not sure how to solve this one. Any help would be appreciated.

• It is surely a typo: the second entry should be $x_{2}^{n-1}$. So now it's pretty clear how to procced Mar 19 at 7:44
• The "big picture" for this type of determinants is Schur polynomials but maybe, in this case it's a sledgehammer for this little fly... Mar 19 at 7:48
• Thanks! Yes I was doubtful this must be a typo. Also yes, while it might be a sledgehammer for this one, I was able to figure out the solution using ideas from Schur polynomials. Mar 20 at 3:11

Preliminaries.

Let $$D$$ denote the diagonal matrix whose entries are $$x_1, x_2, \ldots$$. Note that the characteristic polynomial of $$D$$ is $$p_D(\lambda ) =\Pi_j (\lambda - x_j) = \lambda^n +\sum_{j=1}^{n} (-1)^{j} \sigma_j \lambda^{n-j}$$ where the $$\sigma_j$$ are the elementary symmetric polynomials in these values. In particular $$\sigma_1=tr(D)=\sum_j x_j$$.

The Cayley-Hamilton Theorem asserts that $$p_D(D)=0$$. Thus one obtains the recurrence relation $$D^n = tr(D) D^{n-1}+\ldots$$ where the suppressed terms are lower powers of $$D$$. This recurrence relation is the key identity needed below.

Problem solution

Returning to the problem you posted, let $$L$$ denote the matrix on the left and let $$M$$ denote the standard vanderMonde matrix that appears on the right.

(1) The matrix $$M$$ can be generated as follows. Its first row is $$v_0=[1,1,\ldots,1]$$.

Multiplying the initial row vector $$v_0$$ by successive powers of $$D$$ generates the row vectors $$v_{k+1}= Dv_k = D^{k} v_0$$. That is, $$M$$ is generated by powers $$k=0,1,2 \ldots n-1$$.

Write the list of rows of $$M$$ as $$\{ v_0, v_1, \ldots, v_{n-1}\}$$

(2) In contrast $$L$$ consists of the rows $$\{v_0, v_1 , \ldots v_{n-2}, v_n\}$$

But from the recurrence relation derived above, the row entry $$v_{n}$$ in $$L$$ can be replaced by $$\sigma_1 v_{n-1}$$ plus a linear combination of other rows in $$L$$ that have no affect when we expand the determinant. After this replacement we obtain a matrix that looks like $$M$$ except for the extra factor $$\sigma_1$$. Thus $$det(L)= \sigma_1 det (M)$$.

• Thanks @MathWonk! However isn't $tr(D) = \sum_{j}{x_j^j}$ ? Mar 20 at 2:41
• Nope. Recall that $D$ is the diagonal matrix, not M or N. Mar 20 at 16:00