Problem 2 - Determinants

Let $a_1, a_2, \dots, a_n$ be $n$ real numbers, and $A$ be an $n \times n$ matrix defined by

$$A = \left( \begin{tabular}{cccc} 1 & 1 & $\cdots$ & 1 \\ $a_1$ ... ... $a_1^{n-1}$ & $a_2^{n-1}$ & $\cdots$ & $a_n^{n-1}$ \\ \end{tabular} \right)$$

1.

Compute the determinant of $A$.

2.

Give a necessary and sufficient condition that there exists $\boldsymbol{x} \neq \boldsymbol{0}$ satisfying $A \boldsymbol{x} = \boldsymbol{0}$.

1.

We shall use the following polynomial:

$$\Pi_{i=1}^{n-1}(x-a_i) = x^{n-1} + \sum_{k=1}^{n-2}\lambda_k x^k$$

Without changing the value of the determinant, we can make the following transformation:

$$\det(A) = \left\vert \begin{tabular}{cccc} 1 & 1 & $\cdots$ & 1 \... ... & $a_n^{n-1} + \sum_{k=1}^{n-2}\lambda_k a_n^k$ \\ \end{tabular} \right\vert$$

that we can also write as follows:

$$\det(A) = \left\vert \begin{tabular}{cccc} 1 & 1 & $\cdots$ & 1 \... ..._2-a_i)$ & $\cdots$ & $\Pi_{i=1}^{n-1}(a_n-a_i)$ \\ \end{tabular} \right\vert$$

On the last line, the first $n-1$ elements are 0 and we are in fact given the following determinant:

$$\det(A) = \left\vert \begin{tabular}{cccc} 1 & 1 & $\cdots$ & 1 \... ...0$ & $0$ & $\cdots$ & $\Pi_{i=1}^{n-1}(a_n-a_i)$ \\ \end{tabular} \right\vert$$

We can develop the determinant according to the last line:

$$\det(A) = \det(A_{n-1})\Pi_{i=1}^{n-1}(a_n - a_i)$$

where $A_{n-1}$ is the same matrix as $A$ but without the last line and the rightmost column. We can operate the same transformations on $A_{n-1}$ leading to:

$$\det(A) = \det(A_{n-2})\Pi_{i=1}^{n-2}(a_{n-1} - a_i)\Pi_{i=1}^{n-1}(a_n - a_i)$$

Inductively:

$$\det(A) = (a_2-a_1)\Pi_{i=1}^2(a_3-a_i) \cdots \Pi_{i=1}^{n-2}(a_{n-1} - a_i)\Pi_{i=1}^{n-1}(a_n - a_i)$$

that we can also written:

$$\det(A) = \Pi_{1\leq i<j\leq n}(a_j-a_i)$$

2.

$Ax=0$ with $x\neq 0$ means the linear application that $A$ represents is not a one-to-one mapping and thus $\det(A)=0$ which occurs iff $\exists i,j, i\neq j, x_i = x_j$.