Abstract Nonsense

Crushing one theorem at a time

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^{*}

which, just adding entrywise, gives

\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

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October 4, 2010 - Posted by | Computations, Fun Problems, Linear Algebra | , ,

2 Comments »

  1. thanks alex!

    Comment by Yunjiang Jiang | March 31, 2011 | Reply

    • Of course, friend!

      Comment by drexel28 | March 31, 2011 | Reply


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