Determinant

Multiplicative Property¶

In this section we assume the usual definition of a determinant, from which one can derive the following properties that we will need:

Now our task is to find all functions $f(A)$ from matrices to real numbers that satisfy the multiplicative property:

We will use (4) repeatedly in the following calculation. First we decompose the matrix $A=HU$ using a polar decomposition, where $H$ is Hermitian and $U$ is unitary. Both $H$ and $U$ are individually diagonalizable (but not mutually diagonalizable, since $A$ is not in general diagonalizable). For any matrix $C$ that is diagonalizable, we can write $C=P D P^{-1}$ where $D$ is diagonal and $P$ is invertible. Then:

That means that if a matrix is diagonalizable, then $f$ applied to that matrix is equal to $f$ applied to the diagonal matrix.

Below we will use matrices $P_i$ that are unit matrices with the 1st and $i$-th rows permuted. Now we can compute, using the fact that $H$ and $U$ can be decomposed into diagonal matrices $D_H$ and $D_U$, and their product is also a diagonal matrix $D=D_H D_U$:

where we defined a function $g(x)$ using

Let’s compute the functional equation for $g(x)$:

In the above computations we have only used (4) and no other assumptions.

Let’s summarize our results. From (4) it follows that:

for numbers $x$, $y$.

The equation (22) can be solved in several different ways.

Approach 1: Assuming that $f(A)$ (and thus $g(x)$) is measurable, the multiplicative Cauchy equation is solved in Homomorphisms $\mathbb C^*\to\mathbb C^*$. For real-valued functions this gives $g(x)=0$ or, on $\mathbb R^*$, $g(x)=\operatorname{sgn}(x)^\epsilon |x|^s$ with $\epsilon\in\{0,1\}$ and $s\in\mathbb R$. Consequently the only nonzero solution to (4) is a power of a determinant (optionally multiplied by the signum function). Without measurability, there are also highly pathological solutions.

Approach 2: Instead of assuming measurability, we can instead assume homogeneity:

where $n$ is the (integer) dimension of the matrix $A$. Using (21) and (22) we get:

Comparing the LHS and RHS we can see that $g(\lambda)=\lambda$, or:

and from (21) we get:

Summary:

• Assuming equations (4) and (23) we obtained (27) as the only nonzero solution. That means that we can define a determinant using (4) and (23) and no other assumptions.

• Alternatively, if we only assume (4) and that $f(A)$ is measurable, then the only nonzero solution is a power of a determinant (optionally multiplied with the signum function).

Cauchy Functional Equations¶

The additive and multiplicative Cauchy functional equations, including the measurable solutions needed above, are collected in Homomorphisms $\mathbb C^*\to\mathbb C^*$.

Geometric Definition¶

The way to define a determinant using area/volume from geometry is as follows:

for $c>0$:

All these formulas can be “derived” from intuitive properties of areas in 2D. Another equivalent set of requirements is:

Let’s use the first set of requirements below. We get:

from which it follows:

Now we derive antisymmetry:

and we get:

Now we derive a formula for $V$:

This proves both existence and uniqueness.

Coordinate Definition¶

Determinant of a 2x2 matrix $(\mathbf{A})_{ij}=a_{ij}$ is defined as:

and for a 3x3 matrix $(\mathbf{A})_{ij}=a_{ij}$:

And so on in any dimension. The symbol $\epsilon^{ijk}$ is a Levi-Civita symbol. All the properties of determinants are encoded in the Levi-Civita symbol, so we need to understand it well, but the rest are just usual tensor manipulations.

Let us now derive the multiplicative formula for determinants:

Let’s show it for 2x2 matrices. The LHS is:

The RHS is:

Now we need to prove $\det\mathbf{A}^T = \det\mathbf{A}$. We can do that using the factorial formula by relabeling indices as follows:

We have thus shown:

Alternative Proof¶

First we need the following identity:

Contracting both sides with $\epsilon^{ij}$ we get:

Note that $\epsilon^{kl} a_{ki} a_{lj} = \epsilon^{kl} a_{ik} a_{jl}$ due to the transposition property derived using the last equation in the previous section:

Now we can compute: