# Abstract Nonsense

## Determinant of an Exponential

Point of post: In this post I prove a neat theorem from matrix analysis three different ways.

There’s a neat little theorem which for some reason, I haven’t heard of until now. I’ll prove it in three different ways. The theorem

Theorem: Let $\mathfrak{M}_{n}$ be the set of all $n\times n$ complex matrices and consider the topology induced by the norm

$\left\|\begin{bmatrix} \zeta_{11} & \cdots & \zeta_{1n}\\ \vdots & \ddots & \vdots \\ \zeta_{n1} & \cdots & \zeta_{nn}\end{bmatrix}\right\|=\sqrt{|\zeta_{11}|^2+\cdots+|\zeta_{nn}|^2}$

Then, given a matrix $A\in\mathfrak{M}_n$ such that

$\displaystyle \sum_{k=0}^{\infty}\frac{A^k}{k!}=\exp\left(A\right)$

converges, the following formula holds:

$\det\exp\left(A\right)=\exp\left(\text{tr}\left(A\right)\right)$

As stated we will prove it three ways.

Proof 1: By Schur’s Theorem we have that

$A=U\begin{bmatrix}\lambda_1 & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \lambda_n\end{bmatrix}U^{*}$

where $\{\lambda_1,\cdots,\lambda_n\}$ is the multiset of eigenvalues, $U$ is some unitary matrix, and $*$ is an arbitrary, inconsequential number. And so,

$A^k=U\begin{bmatrix}\lambda_1^k & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \lambda_n^k\end{bmatrix}U^{*}$

And so, for a fixed $n_0\in\mathbb{N}$

$\displaystyle \sum_{k=0}^{n_0}\frac{A^k}{k!}=\sum_{k=0}^{n_0}\left(U\begin{bmatrix}\frac{\lambda_1^k}{k!} & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \frac{\lambda_n^k}{k!}\end{bmatrix}U^{*}\right)=U\sum_{k=0}^{n_0}\begin{bmatrix}\frac{\lambda_1^k}{k!} & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \frac{\lambda_n^k}{k!}\end{bmatrix}U^{*}$

$\displaystyle U\begin{bmatrix}\sum_{k=0}^{n_0}\frac{\lambda_1^k }{k!}& * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \sum_{k=0}^{n_0}\frac{\lambda_n^k}{k!}\end{bmatrix}U^{*}$

And thus,

$\displaystyle \sum_{k=0}^{\infty}\frac{A^k}{k!}=\lim_{n\to\infty}\sum_{k=0}^{n}\frac{A^k}{k!}=\lim_{n\to\infty}U\begin{bmatrix}\sum_{k=0}^{n}\frac{\lambda_1^k }{k!}& * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \sum_{k=0}^{n}\frac{\lambda_n^k}{k!}\end{bmatrix}U^{*}$

but recalling that we’re treating this matrix as really being some $n^2$-tuple, and we assumed that the limit exists we see that (also remembering that the matrices $U,U^{*}$ are “constants”) the above equals

$\displaystyle U\begin{bmatrix}\lim_{n\to\infty}\sum_{k=0}^{n}\frac{\lambda_1^k }{k!}& * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & \lim_{n\to\infty}\sum_{k=0}^{n}\frac{\lambda_n^k}{k!}\end{bmatrix}U^{*}$

where we’ve disregarded that each of the $*$‘s is evaluated at the limit, and technically is a “different” $*$, but it turns out it won’t matter (reusing the asterisk for different values is a common occurrences in matrix analysis textbooks). But, evidently we see that the above evaluates to

$U\begin{bmatrix}e^{\lambda_1} & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & e^{\lambda_n}\end{bmatrix}U^{*}$

And so

$\det\left(\text{exp}\left(A\right)\right)=\det\left(U\begin{bmatrix}e^{\lambda_1} & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & e^{\lambda_n}\end{bmatrix}U^{*}\right)=\det\left(U\right)\det\left(\begin{bmatrix}e^{\lambda_1} & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & e^{\lambda_n}\end{bmatrix}\right)\det\left(U^{*}\right)$

But,

$\det(U)\det\left(U^{*}\right)=\det\left(UU^{*}\right)=\det\left(I_n\right)=1$

and thus

$\det\left(\text{exp}\left(A\right)\right)=\det\left(\begin{bmatrix}e^{\lambda_1} & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & *\\ 0 & \cdots & 0 & e^{\lambda_n}\end{bmatrix}\right)$

But, the determinant of a triangular matrix is the product of it’s diagonal elements. So, finally

$\det\left(\text{exp}\left(A\right)\right)=e^{\lambda_1}\cdots e^{\lambda_n}=\text{exp}\left(\lambda_1+\cdots+\lambda_n\right)=\text{exp}\left(\text{tr}\left(A\right)\right)$

$\blacksquare$

Proof 2: To cut the length of this already long post I will assume the reader is able to prove the following lemma

Lemma: Let $A\in\mathfrak{M}_n$ then if $\{\lambda_1,\cdots,\lambda_n\}$ is the multiset of eigenvalues of $A$, then for any polynomial $P(\cdot)$ we have that $\left\{P\left(\lambda_1\right),\cdots,P\left(\lambda_n\right)\right\}$ is the multiset of eigenvalues for $P(A)$

Proof: Omitted. $\blacksquare$

Now, from this we see that if $A$ has the multiset of eigenvalues of $\{\lambda_1,\cdots,\lambda_n\}$ then for a fixed $n_0\in\mathbb{N}$ we see that the multiset of eigenvalues for

$\displaystyle \sum_{k=0}^{n_0}\frac{A^k}{k!}$

is

$\displaystyle \left\{\sum_{k=0}^{n_0}\frac{\lambda_1^k}{k!},\cdots,\sum_{k=0}^{n_0}\frac{\lambda_n^k}{k!}\right\}$

and so (since the determinant is the product of the eigenvalues of matrix counting multiplicity)

$\displaystyle \det\sum_{k=0}^{n_0}\frac{A^k}{k!}=\prod_{j=1}^{n}\sum_{k=0}^{n_0}\frac{\lambda_j^k}{k!}$

Thus,

$\displaystyle \det\left(\text{exp}\left(A\right)\right)=\det\left(\sum_{k=0}^{\infty}\frac{A^k}{k!}\right)=\det\left(\lim_{m\to\infty}\sum_{k=0}^{m}\frac{A^k}{k!}\right)$

But, notice that since

$\det:\mathfrak{M}_n\to\mathbb{C}$

is continuous (it’s a polynomial of the $n^2$-tuple’s coordinates) we see that

$\displaystyle \det\left(\text{exp}\left(A\right)\right)=\det\left(\lim_{m\to\infty}\sum_{k=0}^{m}\frac{A^k}{k!}\right)=\lim_{m\to\infty}\det\left(\sum_{k=0}^{m}\frac{A^k}{k!}\right)$

and so by prior discussion

$\displaystyle \det\left(\text{exp}\left(A\right)\right)=\lim_{m\to\infty}\det\left(\sum_{k=0}^{m}\frac{A^k}{k!}\right)=\lim_{m\to\infty}\prod_{j=1}^{n}\sum_{k=0}^{m}\frac{\lambda_j^k}{k!}$

But, since each of the factors in the product converges we may split the limit across multiplication to finally get

$\displaystyle \det\left(\text{exp}\left(A\right)\right)=\prod_{j=1}^{n}\lim_{m\to\infty}\sum_{k=0}^{m}\frac{\lambda_{j}^k}{k!}=\prod_{j=1}^{n}e^{\lambda_j}=e^{\lambda_1+\cdots+\lambda_n}=\text{exp}\left(\text{tr}\left(A\right)\right)$

as desired. $\blacksquare$

Proof 3: This is similar to Proof 1 except it doesn’t need a theorem as powerful as Schur’s, but the theorem I’m going to use might actually be a consequence…so haha I don’t know if there’s a point. Oh well..

Lemma: Let $\mathfrak{M}_n$ have the topology described above, then the set of all diagonalizable matrices $\mathcal{D}$ is dense in $\mathfrak{M}_n$.

Proof: Omitted. $\blacksquare$

Now, let $A\in\mathcal{D}$, then if $\{\lambda_1,\cdots,\lambda_n\}$ is the multiset of eigenvalues for $A$ we have that

$A=S\text{ diag}(\lambda_1,\cdots,\lambda_n)S^{-1}$

for some $S\in\mathfrak{M}_n$. Thus,

$A^k=S\text{ diag}\left(\lambda_1^k,\cdots,\lambda_n^k\right)S^{-1}$

And, so more simply than before, for a fixed $n_0\in\mathbb{N}$ we have that

$\displaystyle \sum_{k=0}^{n_0}\frac{A^k}{k!}=S\text{ diag}\left(\sum_{k=0}^{n_0}\frac{\lambda_1^k}{k!},\cdots,\sum_{k=0}^{n_0}\frac{\lambda_n^k}{k!}\right)S^{-1}$

And so

$\displaystyle \text{exp}\left(A\right)=\sum_{k=0}^{\infty}\frac{A^k}{k!}=\lim_{m\to\infty}\sum_{k=0}^{m}\frac{A^k}{k!}=\lim_{m\to\infty}S\text{ diag}\left(\sum_{k=0}^{m}\frac{\lambda_1^k}{k!},\cdots,\sum_{k=0}^{m}\frac{\lambda_n^k}{k!}\right)S^{-1}$

But, as was said before (or similarly) the $S,S^{-1}$ are “constant” and so by the way we are defining the topology we may pass the limit to the entries of the diagonal matrix. Namely, the above says

$\displaystyle \text{exp}\left(A\right)=S\text{ diag}\left(\lim_{m\to\infty}\sum_{k=0}^{m}\frac{\lambda_1^k}{k!},\cdots,\lim_{m\to\infty}\sum_{k=0}^{m}\frac{\lambda_n^k}{k!}\right)S^{-1}=S\text{ diag}\left(e^{\lambda_1},\cdots,e^{\lambda_n}\right)S^{-1}$

Thus,

$\det\left(\text{exp}\left(A\right)\right)=\det\left(S\text{ diag}\left(e^{\lambda_1},\cdots,e^{\lambda_n}\right)S^{-1}\right)=\text{exp}\left(\text{tr}\left(A\right)\right)$

(where we’ve implicitly used that $\det(S)\det(S^{-1})=\det(SS^{-1})=\det(I_n)=1$). It follows that

$\text{det}:\mathfrak{M}_n\to\mathbb{C}:A\mapsto\text{exp}\left(A\right)$

and

$\text{exp}\circ\text{tr}:\mathfrak{M}_n\to\mathbb{C}:A\mapsto \text{exp}\left(\text{tr}\left(A\right)\right)$

agree on a dense subset of $\mathfrak{M}_n$, but since both are evidently continuous it is a simple topological consequence that they must agree on all of $\mathfrak{M}_n$ from where the conclusion follows. $\blacksquare$