Understanding the determinant of an infinite matrix 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:


*

*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?

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

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

 A: 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.
