# Compare determinants of matrices with different dimensions

How valid is to compare the determinants of matrices with different dimensions? e.g. compare a determinant $D1$ derived from a $N\times N$ matrix with the determinant $D2$ derived from a $M\times M$ matrix.

I've read that the determinant represents the volume of the $N-$dimensional object that is defined by the elements of the matrix.

If this is correct then comparing volumes of different "objects" doesn't sound as incorrect. But having in mind that these "objects" came from two different spaces (a $N-$dimensional and a $M-$dimensional) how valid is that?

• This might be interesting when you compare the determinant of matrix and the determinant of one its submatrices (by deleting some rows and columns). – Surb May 6 '15 at 8:22
• you should try to make your question more precise because it is not clear what exactly you are asking – user126154 May 6 '15 at 8:45
• So, here is the concept behind my question. I have a bunch of clusters. Each cluster has different number of elements. e.g. one cluster has N items, another one has M, the other has L and so on. By calculating the distances between elements of a cluster, I end with a bunch of different matrices (a NxN, MxM, LxL and so on). Then, I can get the determinant of each one, but how valid is to compare them? Obviously, I need to find the one that has the minimum determinant (volume). – maus May 6 '15 at 8:53

Consider an identity or an inequality of the following form:

$$\tag{*} \det A_n=C\cdot \det B_m, \qquad [\text{or }\le,\ \text{or }\ge], \qquad \forall A_n\in \mathcal{A},\ \forall B_m\in \mathcal{B}.$$

Here $C$ is a constant and $\mathcal{A}, \mathcal{B}$ are some families of $n\times n$ and $m\times m$ matrices.

If $n\ne m$ then the relation $(*)$ has a serious chance of being incorrect and should be regarded with suspicion if it arises in actual computations. Namely, if $\mathcal{A}$ and $\mathcal{B}$ are closed under scaling, that is

$$(A\in \mathcal{A},\ B\in \mathcal{B},\ \lambda\ge 0)\ \Rightarrow\ \lambda A\in \mathcal{A}\ \text{and}\ \lambda B \in \mathcal{B},$$

then $(*)$ is incorrect (except for trivial cases, such as having $C=0$ and the like). Indeed, if $(*)$ holds, then one should have

$$\lambda^n\det A_n = C\cdot \lambda^m\det B_m, \qquad [\text{or }\le,\ \text{or }\ge],$$

which forces a contradiction in the limit $\lambda \to 0$ or $\lambda \to \infty$ (except the trivial case of an identity $0=0$, of course).

This is usually called scaling argument.