Let $x_1,..., x_k$ be vectors in $\mathbb{R}^n$. Show that $V(x_1 , ... ,\lambda x_i, ... ,x_k) = |\lambda|V(x_1, ... ,x_k)$

Question

Let $x_1,..., x_k$ be vectors in $\mathbb{R}^n$. Show that $$V(x_1 , ... ,\lambda x_i, ... ,x_k) = |\lambda|V(x_1, ... ,x_k)$$

I know that $V(x_1, ... ,x_k)=\{\det[(x_1, ... ,x_k)^{tr}(x_1, ... ,x_k)]\}^{1/2}$ but I do not know what else to do, could someone tell me what else I could do? I tried to call $x_i=\begin{bmatrix}x_{i1}\\. \\. \\. \\x_{in} \end{bmatrix}$ but I find it a bit cumbersome to calculate this by expanding everything, is there another way?

Answer 1

You can write $$V(x_1,\ldots, x_n) = \sqrt{\det \big((x_i\cdot x_j)_{i,j=1}^n \big)},$$where $\cdot$ denotes dot product. When you consider $V(x_1,\ldots, \lambda x_k,\ldots, x_n)$, the $k$-th line and the $k$-th column of the determinant are multiplied by $\lambda$. These $\lambda$'s gets factored out of the determinant, and the absolute values distributes. Hence $$V(x_1,\ldots, \lambda x_k,\ldots, x_n) = \sqrt{\lambda^2 \det((x_i\cdot x_j)_{i,j=1}^n)} = |\lambda|\sqrt{\det((x_i\cdot x_j)_{i,j=1}^n)} = |\lambda| V(x_1,\ldots, x_n).$$