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For the 2x2 matrix case, the determinant of $1+A$ is

$\det(1+A) = 1 + \mathop{tr} A + \det A$

so here the criterion $\det(1+A)\neq0$ can be reformulated in terms of $A$'s trace and determinant, but is there a generalization of this for arbitrary matrices and linear operators?

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    Yes, this is only a sufficient condition. The point is that this criterion also works in infinite dimensions (Banach spaces). I don't know a necessary condition, though (and think it's unlikely to exist, even for Hilbert spaces).2011-08-29

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If $\lambda$ is an eigenvalue of $A$, then $1+\lambda$ is an eigenvalue of $I+A$. Indeed, let $v$ be an eigenvector to the eigenvalue $\lambda$, i.e. $Av = \lambda v$. Then $(I+A)v = v + \lambda v = (1+\lambda)v$.

But then $\det(I+A) = \prod_{i} (1+\lambda_i)$, where the product is over all eigenvalues $\lambda_i$ of $A$ (and the same factors can turn up multiple times).

So $I+A$ is invertible if and only if $\lambda_i \ne -1$ for all $\lambda_i$.

Note that the above is the same as your criterion for $2\times 2$ matrices:

$\prod_{i} (1+\lambda_i) = 1 + (\lambda_1 + \lambda_2) + \lambda_1 \lambda_2 = 1 + \mathrm{tr} A + \det A$

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    @Tobias: Maybe you can make use of some ideas on the [Fredholm determinant](http://en.wikipedia.org/wiki/Fredholm_determinant), but that's just a shot in the dark.2011-08-29