[Math] Understanding the determinant of an infinite matrix

determinantfunctional-analysislinear algebramatrices

The fredholm determinant of an infinite matrix $A$ is defined by
$$
\det(I+A) = \exp(\text{tr}(\log(I+A))).
$$
I'm trying to understand the formula and how to use it.

To be concrete here are my questions:

  1. Let
    $$
    A=I=\begin{pmatrix}1 & 0 & 0 & \ldots \\
    0 & 1 & 0 & \ddots \\
    \vdots & \ddots & \ddots &\ddots \\
    \end{pmatrix}
    $$
    ($A$ is the infinite matrix with diagonal elements equal to 1 and 0's everywhere else). If $A$ were finite dimensional then mutlipyling the diagonal elements, the determinant equals 1. It seems natural that the infinite matrix should also have determinant equal to 1 but I don't see how the above formula gets this.
    What about a triangular matrix with diagonal elements equal to 1?
    What about a diagonal matrix with elements not equal to 1?

  2. Why is the definition not $\det(A) = \exp(\text{tr}(\log(A)))$?

  3. Say $\det(I+A)=3$ then what is $\det(A)$?

Best Answer

Math with infinite matrices is tricky, as is anything to do with infinite-dimensional linear algebra. Note in particular that the Fredholm determinant $\det(I + A)$ will only be defined if $A$ is a trace-class matrix, i.e. an operator for which $\sqrt{A^TA}$ has finite trace. So to answer your last two questions:

Regarding 2: if we wanted to define $\det(A)$ in the same way, then we would only be able to use the trick that defines the Fredholm derivative if $A - I$ is trace-class. It's easier to talk about trace-class operators than it is to talk about "the set of operators $A$ for which $A - I$ is trace-class", hence the definition given.

Regarding 3: if $\det(I + A) = 3$, then $\det(A)$ is not defined. In particular, if $A$ is a trace-class operator, then $A + \alpha I$ will only be trace-class if $\alpha = 0$.

Regarding the first question: to calculate $\det(I)$, we can plug in $A = 0$ into the definition to get $$ \det(I + 0) = \exp(\operatorname{tr}(\log(I+ 0))). $$ We have $\log(I) = 0$ (the zero-matrix), so the trace of the log is zero, so the determinant is $\exp(0) = 1$.

A triangular matrix might not be trace class. For instance, as long as we have an infinite diagonal that doesn't converge to zero, the matrix fails to be trace-class. In the same way, for some diagonal matrices $A$ it is not true that $A - I$ is trace class. Whenever $A - I$ is not trace class, $\det(A)$ is not defined in this context.

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