# Linear transformation of Gram determinant

The Gram determinant of the vectors $$\mathbf x_1,\mathbf x_2,\dots,\mathbf x_k \in \mathbb R^n$$ is defined as:

$$\Gamma (\mathbf{x}_1, \ldots ., \mathbf{x}_k) = \left|\begin{array}{ccc} \mathbf{x}_1^{\top} \mathbf{x}_1 & \cdots & \mathbf{x}_1^{\top} \mathbf{x}_k\\ \vdots & \ddots & \vdots\\ \mathbf{x}_k^{\top} \mathbf{x}_1 & \cdots & \mathbf{x}_k^{\top} \mathbf{x}_k \end{array}\right| = \det (\mathbf{X}^{\top} \mathbf{X}) \qquad\qquad (1)$$

where $$\mathbf{X}$$ is the $$n \times k$$ rectangular matrix with columns $$\mathbf{x}_1, \ldots, \mathbf{x}_k$$. It gives the "sub-dimensional" squared volume of the parallelepiped spanned by these vectors.

If I apply a linear transformation $$\mathbf A\in\mathbb{R}^{n\times n}$$ on the vectors $$\mathbf x_i$$, the volume changes to

$$\Gamma (\mathbf{A}\mathbf{x}_1, \ldots, \mathbf{A}\mathbf{x}_k) = \det (\mathbf{X}^{\top} \mathbf{A}^{\top} \mathbf{A}\mathbf{X}) \qquad\qquad (2)$$

although I see no obvious way to simplify this. I'd expect the volume to change in a simple manner after a linear transformation, like a constant volume dilation, but

$$\Gamma (\mathbf{A}\mathbf{x}_1, \ldots ., \mathbf{A}\mathbf{x}_k) \ne [\det (\mathbf{A})]^2 \Gamma(\mathbf{x}_1, \ldots, \mathbf{x}_k)$$

in general (unless $$k=n$$ or $$\mathbf B$$ is proportional to the identity matrix).

I could apply a $$k\times k$$ matrix $$\mathbf B$$ "on the right" and use the identity,

$$\det (\mathbf{B}^{\top} \mathbf{X}^{\top} \mathbf{X}\mathbf{B}) = \det (\mathbf{X}^{\top} \mathbf{X}) [\det (\mathbf{B})]^2 \qquad\qquad (3)$$

i.e., a volume dilation. But of course this is not the same as (2) above. Here the matrix $$\mathbf B$$ is acting on the rows of $$\mathbf X$$ and the geometrical meaning of this operation is not obvious to me.

Why my intuition is fails here? Why a linear transformation does not result in a constant volume dilation of sub-dimensional volumes? And what is the geometrical meaning of (3)?

I am looking for answers to these questions that offer some intuition, and particularly geometrical insight.

First, the equation for the volume should be $$(\det(X^{\top} X))^{1/2}$$.
Consider for instance the transformation of $$\mathbb{R}^4=\mathbb{R}^2\times\mathbb{R}^2$$ given by $$\mathrm{diag}(2,2,3,3)$$. If you consider a parallelogram in the first $$\mathbb{R}^2$$, the transformation multiplies the 2-volume (area) of the parallelogram with a factor $$2\cdot 2=4$$. A parallelogram in the second $$\mathbb{R}^2$$ would increase with a factor $$3\cdot 3=9$$.
If the parallelopipedum is a $$n$$-volume (so $$k=n$$), then a linear transformation just multiplies the volume with a factor $$\det A$$. To put it in a very non-mathematical, but intuitive way: it doesn't matter how you orient it, a volume will always stretch/shrink in all directions.
I can't give an interpretation of equation (3), I don't find a realistic interpretation of $$\mathbf{X}\mathbf{B}$$ that's related to the parallelopipedum.