Bound on determinant of matrix exponential

Question

Given bounds on the value of the determinant of an infinite dimensional square matrix $A$, what are the bounds on the determinant of the matrix exponential of that matrix?

Answer 1

We are asked to provide a bound on the determinant of $e^A$ in terms of the determinant of square matrix $A$, possibly allowing for the size $n$ to be used in formulating the bound in a manner that allows us to take a limit as $n\to \infty$.

To show that no bound of the requested form is possible, it suffices to construct a family of matrices $A_n$ of respective sizes $n\times n$ for all $n\ge 2$ such that $\det A_n = 0$ but $\det e^{A_n}$ is arbitrarily large.

Let $A_n$ be the $n\times n$ matrix whose entries are all equal to $a$. Then for $n\ge 2$ we have duplicate rows and $\det A_n = 0$. However the determinant of the matrix exponential, as @BillCook explains, is the exponential of the trace of the matrix. In this case:

$$ \det e^{A_n} = e^{\operatorname{tr} A_n} = e^{na} $$

The requested bound would entail a sequence of real functions $\Phi_n$ so that for all sufficiently large $n$:

$$ \det e^{A_n} \le \Phi_n(\det A_n) $$

But we cannot provide such a function for any $n\ge 2$, much less for all sufficiently large $n$, because with the specific matrices $A_n$ it would mean:

$$ e^{na} \le \Phi_n(0) $$

However large the value $\Phi_n(0)$ is, we can always choose $a$ in a way that causes such an inequality to fail. For example, $a = |\Phi_n(0)|$ would cause such a failure.

Answer 2

$$\mathrm{det}(e^A) = e^{\mathrm{tr}(A)}$$

So the determinant is the exponential of the trace. So you probably won't find many "bounds" because the actual value is pretty easy to compute.

However, you could say things like: Let $\lambda_{\mathrm{max}}$ be the eigenvalue of maximum modulus (i.e. absolute value when all are real). Then $\mathrm{tr}(A)$ is the sum of the eigenvalues (counting multiplicity) and so $|\mathrm{tr}(A)| \leq n|\lambda_{\mathrm{max}}|$ when $A$ is $n \times n$. Thus $\mathrm{det}(A) \leq e^{n|\lambda_{\mathrm{max}}|}$.