Determinant-Trace-Equation using Exterior Algebra

This post assumes knowledge of the basics of exterior algebra.

This is a short proof of the fact that, for any linear map $𝐹 : ℝ^{𝑛} \to ℝ^{𝑛}$ , the equality

det exp 𝐹 = exp tr 𝐹

holds.

I was inspired to write this after seeing a 17-minute YouTube video where this equation was proven using matrix algebra. Noting how determinants are so much easier to define and manipulate in geometric algebra, or even exterior algebra (since we won’t need the inner product here), I set out to find a shorter proof.

Preliminaries

Let’s look at some properties we’ll need.

The operator exponential $exp (𝑡 𝐹)$ has the derivative
$\frac{𝑑}{𝑑 𝑡} exp (𝑡 𝐹) (-) = 𝐹 (exp (𝑡 𝐹) (-)) .$
The scalar differential equation $𝑓^{'} (𝑡) = 𝑐 𝑓 (𝑡)$ with initial condition $𝑓 (0) = 1$ has the unique solution
$𝑓 (𝑡) = exp (𝑐 𝑡) .$
The trace of $𝐹$ is given by, for any basis ${𝑎_{𝑖}}_{𝑖 = 1}^{𝑛}$ of $ℝ^{𝑛}$ ,
$tr 𝐹 = \sum_{𝑖 = 1}^{𝑛} 𝑎^{* 𝑖} (𝐹 (𝑎_{𝑖}))$
where ${𝑎^{* 𝑖}}_{𝑖 = 1}^{𝑛}$ is the dual basis to ${𝑎_{𝑖}}_{𝑖 = 1}^{𝑛}$ , i.e. $𝑎^{* 𝑖} \in 𝑉^{*}$ and
$𝑎^{* 𝑖} (𝑎_{𝑗}) = 𝛿_{𝑗}^{𝑖} = {\begin{cases} 1, & 𝑖 = 𝑗 \\ 0, & 𝑖 \neq 𝑗 . \end{cases}$
(This is where the exterior algebra comes in.) The determinant of $𝐹$ is given by the formula
$⋀_{𝑖 = 1}^{𝑛} 𝐹 (𝑎_{𝑖}) = det (𝐹) ⋀_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} .$
This formula is a rigorous version of the notion that “the determinant measures how volume is scaled.”
The derivative of an exterior product follows a product rule:
$\frac{𝑑}{𝑑 𝑡} (⋀_{𝑖 = 1}^{𝑚} 𝑣_{𝑖} (𝑡)) = \sum_{𝑖 = 1}^{𝑚} (⋀_{𝑗 = 1}^{𝑖 - 1} 𝑣_{𝑗} (𝑡)) \land 𝑣_{𝑖}^{'} (𝑡) \land (⋀_{𝑗 = 𝑖 + 1}^{𝑚} 𝑣_{𝑗} (𝑡)) .$
$𝐺 (𝑡) = exp (𝑡 𝐹)$ is invertible, its inverse simply being $𝐺 (- 𝑡)$ , thus, for any basis ${𝑎_{𝑖}}_{𝑖 = 1}^{𝑛}$ , ${𝐺 (𝑡) (𝑎_{𝑖})}_{𝑖 = 1}^{𝑛}$ is also a basis.

I think all of these count as background knowledge and not part of the proof, so I won’t prove them here. In fact, if I were to prove them, I’d have to prove them using other facts, starting a spiral that would only end with me writing a whole textbook.

The proof

The main body of the proof is a big chain equality proving that

\frac{𝑑}{𝑑 𝑡} ⋀_{𝑖 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})) \overset{(*)}{=} tr (𝐹) ⋀_{𝑖 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})),

which, using the above definition of the determinant, can be rewritten to

\frac{𝑑}{𝑑 𝑡} [det (exp (𝑡 𝐹)) ⋀_{𝑖 = 1}^{𝑛} 𝑎_{𝑖}] = tr (𝐹) det (exp (𝑡 𝐹)) ⋀_{𝑖 = 1}^{𝑛} 𝑎_{𝑖},

which then just reduces to

\frac{𝑑}{𝑑 𝑡} det (exp (𝑡 𝐹)) = tr (𝐹) det (exp (𝑡 𝐹)),

which, substituting $𝑡 = 0$ to see that $det (exp (𝑡 𝐹)) |_{𝑡 = 0} = det (id) = 1$ , immediately implies that

det (exp (𝑡 𝐹)) = exp (𝑡 tr (𝐹)) .

(compare fact 2 above)
Substituting $𝑡 = 1$ finally results in the desired equality:

det (exp (𝐹)) = exp (tr (𝐹)) .

Now all that remains is to show equation $(*)$ . For that, first apply the product rule to get

\begin{aligned} \frac{𝑑}{𝑑 𝑡} ⋀_{𝑖 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})) & = \sum_{𝑖 = 1}^{𝑛} (⋀_{𝑗 = 1}^{𝑖 - 1} exp (𝑡 𝐹) (𝑎_{𝑗})) \land \frac{𝑑}{𝑑 𝑡} (exp (𝑡 𝐹) (𝑎_{𝑖})) \land (⋀_{𝑗 = 𝑖 + 1}^{𝑛} exp (𝑡 𝐹) (𝑎_{𝑗})) \\ = \sum_{𝑖 = 1}^{𝑛} (⋀_{𝑗 = 1}^{𝑖 - 1} exp (𝑡 𝐹) (𝑎_{𝑗})) \land 𝐹 (exp (𝑡 𝐹) (𝑎_{𝑖})) \land (⋀_{𝑗 = 𝑖 + 1}^{𝑛} exp (𝑡 𝐹) (𝑎_{𝑗})) . \end{aligned}

Now define ${\tilde{𝑎}}_{𝑗} ≔ exp (𝑡 𝐹) (𝑎_{𝑗})$ for all $𝑗 \in {1, \dots, 𝑛}$ , so that ${{\tilde{𝑎}}_{𝑗}}_{𝑗 = 1}^{𝑛}$ forms a basis (compare the final fact above). Thus, we can decompose $𝐹 ({\tilde{𝑎}}_{𝑖})$ into a linear combination

𝐹 ({\tilde{𝑎}}_{𝑖}) = \sum_{𝑘 = 1}^{𝑛} {\tilde{𝑎}}^{* 𝑘} (𝐹 ({\tilde{𝑎}}_{𝑖})) {\tilde{𝑎}}_{𝑘} .

Substitute that in to obtain

\begin{aligned} \frac{𝑑}{𝑑 𝑡} ⋀_{𝑖 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})) & = \sum_{𝑖 = 1}^{𝑛} (⋀_{𝑗 = 1}^{𝑖 - 1} {\tilde{𝑎}}_{𝑗}) \land (\sum_{𝑘 = 1}^{𝑛} {\tilde{𝑎}}^{* 𝑘} (𝐹 ({\tilde{𝑎}}_{𝑖})) {\tilde{𝑎}}_{𝑘}) \land (⋀_{𝑗 = 𝑖 + 1}^{𝑛} {\tilde{𝑎}}_{𝑗}) \\ = \sum_{𝑘 = 1}^{𝑛} \sum_{𝑖 = 1}^{𝑛} {\tilde{𝑎}}^{* 𝑘} (𝐹 ({\tilde{𝑎}}_{𝑖})) (⋀_{𝑗 = 1}^{𝑖 - 1} {\tilde{𝑎}}_{𝑗}) \land {\tilde{𝑎}}_{𝑘} \land (⋀_{𝑗 = 𝑖 + 1}^{𝑛} {\tilde{𝑎}}_{𝑗}) . \end{aligned}

Now note that the exterior product contains the repeated term ${\tilde{𝑎}}_{𝑘}$ if and only if $𝑘 \neq 𝑖$ , so all those terms vanish, leaving only

\begin{aligned} \frac{𝑑}{𝑑 𝑡} ⋀_{𝑖 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})) & = \sum_{𝑖 = 1}^{𝑛} {\tilde{𝑎}}^{* 𝑖} (𝐹 ({\tilde{𝑎}}_{𝑖})) (⋀_{𝑗 = 1}^{𝑖 - 1} {\tilde{𝑎}}_{𝑗}) \land {\tilde{𝑎}}_{𝑖} \land (⋀_{𝑗 = 𝑖 + 1}^{𝑛} {\tilde{𝑎}}_{𝑗}) \\ = \sum_{𝑖 = 1}^{𝑛} {\tilde{𝑎}}^{* 𝑖} (𝐹 ({\tilde{𝑎}}_{𝑖})) ⋀_{𝑗 = 1}^{𝑛} {\tilde{𝑎}}_{𝑗} \\ = (\sum_{𝑖 = 1}^{𝑛} {\tilde{𝑎}}^{* 𝑖} (𝐹 ({\tilde{𝑎}}_{𝑖}))) (⋀_{𝑗 = 1}^{𝑛} {\tilde{𝑎}}_{𝑗}) . \end{aligned}

Now it is easy to see that the first factor is just the definition of $tr 𝐹$ , so we finally have

\frac{𝑑}{𝑑 𝑡} ⋀_{𝑖 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})) = tr (𝐹) (⋀_{𝑗 = 1}^{𝑛} {\tilde{𝑎}}_{𝑗}) = tr (𝐹) ⋀_{𝑗 = 1}^{𝑛} (exp (𝑡 𝐹) (𝑎_{𝑖})) .

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