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, we show below that the only solutions are $g(x)=0$, $g(x)=|\mathrm{sign}(x)|$, $g(x)=|x|^s$ and $g(x)=\mathrm{sign}(x)|x|^s$. Consequently the only nonzero solution to (4) is a power of a determinant (optionally multiplied with the signum function). If $f(A)$ is not measurable then we also get highly pathological solutions (see the section below). The above is thus the most general solution.

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’s Additive Functional Equation¶

The Cauchy’s functional equation (Wikipedia (2025)) usually means the additive equation of this form:

There are other related equations, such as the Cauchy’s multiplicative functional equation in the next section.

For integer $n$ we get:

By substituting $y=nx$ we get $f(y)=nf(y/n)$, from which $f(y/n)=(1/n)f(y)$. Using these two relations for multiplication and division by an integer we obtain for any rational number $q={p\over q}$, where $p$, $q$ are integers:

We also have to prove it for $p\le 0$ which we can using $f(0)=f(0)+f(0)$ which implies $f(0)=0$ and using $0=f(0)=f(x-x)=f(x)+f(-x)$ from which follows $f(-x)=-f(x)$. Setting $x=1$ we get for all rational $q$:

where $c=f(1)$ is any real constant. If we assume that $f(x)$ is continuous, then this implies that $f(x)=c x$ for all real $x$, since the rational numbers are dense in real numbers. It turns out that one can relax the continuity requirement to only require that $f(x)$ is measurable. Since $f(x)=c x$ is linear, we get:

A linear function $f(x)=ax+b$ can be substituted into (28) and one obtains that $b=0$. So this theorem is a simple way to remember the solutions to the Cauchy’s additive functional equation.

Cauchy’s Multiplicative Functional Equation¶

The Cauchy’s multiplicative functional equation is:

There are only 3 cases that can happen:

• $f(x) = 0\,.$

• $f(x) = 1\,.$

• $f(x) = \begin{cases} h(x) & \text{if } x > 0\,,\\ 0 & \text{if } x = 0\,,\\ \pm h(|x|) & \text{if } x < 0\,.\\ \end{cases}$

In the last case the function $h(x)$ is positive $h(x) > 0$ and defined only for $x>0$. The $x < 0$ case has two options for even and odd $f(x)$. We can equivalently write the two cases explicitly as follows:

• Even: $f(x) = \begin{cases} 0 & \text{if } x = 0\,,\\ h(|x|) & \text{if } x \ne 0\,.\\ \end{cases}$

• Odd: $f(x) = \begin{cases} 0 & \text{if } x = 0\,,\\ \mathrm{sign}(x) h(|x|) & \text{if } x \ne 0\,.\\ \end{cases}$

As a special case for $h(x)=1$ we get:

• Even: $f(x) = |\mathrm{sign}(x)| = \begin{cases} 0 & \text{if } x = 0\,,\\ 1 & \text{if } x \ne 0\,. \end{cases}$

• Odd: $f(x) = \mathrm{sign}(x) = \begin{cases} 0 & \text{if } x = 0\,,\\ 1 & \text{if } x > 0\,,\\ -1 & \text{if } x < 0\,. \end{cases}$

We can then use the substitution $g(x)=\log f(e^x)$ to convert the multiplicative equation into an additive equation from the previous section:

We find the solution using the previous section, and then we compute $h(x)$ using $h(x)=\exp(g(\log x))$, which satisfies:

For measurable $g(x)$ the solution to the additive equation is $g(x) = cx$, so we assume $f(x)$ is measurable (which implies $g(x)$ is measurable) and we get $h(x)=\exp(c\log x)=\exp(\log(x^c))=x^c$ and the four solutions are

• $f(x) = 0\,.$

• $f(x) = 1\,.$

• $f(x) = \begin{cases} 0 & \text{if } x = 0\,,\\ h(|x|) & \text{if } x \ne 0\,,\\ \end{cases} = \begin{cases} 0 & \text{if } x = 0\,,\\ |x|^c & \text{if } x \ne 0\,.\\ \end{cases}$

• $f(x) = \begin{cases} 0 & \text{if } x = 0\,,\\ \mathrm{sign}(x) h(|x|) & \text{if } x \ne 0\,,\\ \end{cases} = \begin{cases} 0 & \text{if } x = 0\,,\\ \mathrm{sign}(x) |x|^c & \text{if } x \ne 0\,.\\ \end{cases}$

The solutions can be equivalently written as:

• $f(x)=0\,.$

• $f(x) = |\mathrm{sign}(x)| = \begin{cases} 0 & \text{if } x = 0\,,\\ 1 & \text{if } x \ne 0\,. \end{cases}$

• $f(x)=|x|^c$, with $f(0)=0$ (needed for $c < 0)$.

• $f(x)=\mathrm{sign}(x)|x|^c$, with $f(0)=0$ (needed for $c < 0)$.

The solution $f(x)=1$ is already included in $|x|^c$ for $c=0$. If $c$ is an integer, then the last two solutions can always be written (for both even and odd integers) as $f(x)=x^c$ and $f(x)=\mathrm{sign}(x) x^c$.

Some examples are: 0, 1, $\mathrm{sign}(x)$, $|\mathrm{sign}(x)|$, $x$, $|x|$, $x^2$, $\mathrm{sign}(x) x^2$, $x^3$, $|x|^3$, $1\over x$, $1\over |x|$,$1\over x^2$,${\mathrm{sign}(x)\over x^2}={|x|\over x^3}$, $1\over x^3$, $\sqrt{|x|}$, $1\over\sqrt{|x|^3}$, etc.

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: