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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.

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First, the equation for the volume should be $(\det(X^{\top} X))^{1/2}$.

The reason why equation (2) holds, is because it really depends on the orientation of the lower-dimensional parallelopided.

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.

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  • $\begingroup$ Thanks, now I understand why sub-dimensional volumes are not all scaled in the same way under generic linear transforms. I would still like to understand (3) better. $\endgroup$ – becko Sep 25 '18 at 12:31

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