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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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Would $\|A\| \lt 1$ or $r(A) \lt 1$ (where $$r(A) = \lim_{n\to\infty} \|A^{n}\|^{1/n}$$ is the spectral radius) be satisfactory for you? Then you can write the inverse as Neumann series $$(1+A)^{-1} = \sum_{n=0}^{\infty} (-1)^{n}A^{n}$$ with convergence on the right hand side. – t.b. Aug 29 '11 at 9:01
@Theo thanks, I'd upvote that as an answer as well, though Sam's seems to be more general (if applicable to general operators as well) – Tobias Kienzler Aug 29 '11 at 9:37
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). – t.b. Aug 29 '11 at 9:44
up vote 5 down vote accepted

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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You can also make use of the induced maps on the exterior algebra, as explained e.g in Qiaochu's answer here. This allows you to avoid speaking explicitly of the eigenvalues (if that is the goal of the question). – t.b. Aug 29 '11 at 9:27
perfect! does this also apply for operators? are there criteria to their spectra? – Tobias Kienzler Aug 29 '11 at 9:36
@Theo: Quite interesting, thanks for the link. If one wanted to avoid eigenvalues, I guess one could also write down the formula in terms of minors of $A$ (which would be very messy, if it works). – Sam Aug 29 '11 at 9:38
@Tobias: It doesn't really help in any way, it just gives a theoretically nice expression and in some sense the elementary symmetric polynomials (linked to in Qiaochu's answer) interpolate between the trace and the determinant. If you want to evaluate the expression, though, it boils down to expanding the determinant of $I+A$ using minors and collecting the terms of the same degree. A big mess unless you diagonalize first (or use the Jordan normal form). – t.b. Aug 29 '11 at 9:49
@Tobias: Maybe you can make use of some ideas on the Fredholm determinant, but that's just a shot in the dark. – t.b. Aug 29 '11 at 10:08

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