How do you compute the determinant of a matrix where the same vector appears as a component of every column?

Question

Assume that every vector $\mathbf{x}_i$ is of length $n$. A textbook I'm reading seems to suggest that the following holds $$ \begin{vmatrix} \rule[-1ex]{0.5pt}{2.5ex} & \rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\ \mathbf{x}_0-\mathbf{x}_1 & \mathbf{x}_0-\mathbf{x}_2 & \cdots & \mathbf{x}_0-\mathbf{x}_n\\ \rule[-1ex]{0.5pt}{2.5ex} & \rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex} \end{vmatrix} = \frac{1}{n!}\: \begin{vmatrix} 1 & 1 & \cdots & 1\\ \mathbf{x}_0 & \mathbf{x}_1 & \cdots & \mathbf{x}_n \end{vmatrix}\,. $$ Is this so? If not, are there any other well-known means to simplify the left-hand side?

Answer 1

$$ \begin{vmatrix} 1 & 1 & \ldots & 1 \\ \mathbf{x_0} & \mathbf{x_1} & \ldots & \mathbf{x_n} \end{vmatrix} = \begin{vmatrix} 1 & 0 & \ldots & 0 \\ \mathbf{x_0} & \mathbf{x_1} - \mathbf{x_0} & \ldots & \mathbf{x_n} - \mathbf{x_0} \end{vmatrix} = \begin{vmatrix} 1 & 0 & \ldots & 0 \\ 0 & \mathbf{x_1} - \mathbf{x_0} & \ldots & \mathbf{x_n} - \mathbf{x_0} \end{vmatrix} $$

which is just

$$ (-1)^n \begin{vmatrix} \mathbf{x_0} - \mathbf{x_1} & \ldots & \mathbf{x_0} - \mathbf{x_n} \end{vmatrix} $$

Answer 2

First evaluate the right hand side: expanding by the first row, we have

$$ \frac{1}{n!} \sum_{i=0}^n (-1)^i \left| \begin{array}{cccccc} | & & | & | & & |\\ \mathbf{x}_0 & \cdots & \mathbf{x}_{i-1} & \mathbf{x}_{i+1} & \cdots & \mathbf{x}_n \\ | & & | & | & & | \end{array} \right| $$

This is the alternating sum of the determinants of all-but-one of the $\mathbf{x}_i$.

Now look at the left-hand-side. Use multilinearity of the determinant. By that I mean the property that $$ \left| \begin{array}{ccccc} | & & | & & | \\ \mathbf{v}_1 & \cdots & \mathbf{v}_i + \mathbf{u}_i & \cdots & \mathbf{v}_n \\ | & & | & & | \end{array} \right| = \left| \begin{array}{ccccc} | && | && | \\ \mathbf{v}_1 & \cdots & \mathbf{v}_i & \cdots & \mathbf{v}_n \\ | && | && | \end{array} \right| + \left| \begin{array}{ccccc} | && | && | \\ \mathbf{v}_1 & \cdots & \mathbf{u}_i & \cdots & \mathbf{v}_n \\ | && | && | \end{array} \right|$$

If you apply this property to each column simultaneously, you see that the left-hand side is the sum over all possibilities where for each column you choose either $\mathbf{x}_0$ or $-\mathbf{x}_i$. But the determinant is also alternating, so any time two columns are the same, the determinant is zero. This means we only get the terms where we choose at most one column to be $\mathbf{x}_0$. So we get

$$ \left| \begin{array}{ccc} | && | \\ -\mathbf{x}_1 & \cdots & -\mathbf{x}_n \\ | && |\end{array} \right| + \sum_{i=1}^n \left| \begin{array}{ccccccc} | && | & | & | && | \\ -\mathbf{x}_1 & \cdots & -\mathbf{x}_{i-1} & \mathbf{x}_0 & -\mathbf{x}_{i+1} & \cdots & -\mathbf{x}_n \\ | && | & | & | && | \end{array} \right| $$

Now if you swap columns so that the $\mathbf{x}_0$ is always the first column, you get the factors of $(-1)^i$. Also, from pulling out the minus signs, each term gets a factor of $(-1)^{n-1}$ or $(-1)^n$.

I might have made a mistake somewhere, but I still don't see where the $n!$ comes from...