I am trying to obtain the determinant of the difference between the identity matrix and an A matrix. The question is such:
det(I-A) or another one is, det(I-AB)
Thanks for the help.
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$\begingroup$ What are you expecting? A magic formula where none exists? $\endgroup$– Michael RenardyCommented Jan 11, 2014 at 12:36
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1$\begingroup$ @MichaelRenardy: I wonder how one can know that a formula does not exist. Anyway, the OP is not a research level question. $\endgroup$– TaladrisCommented Jan 11, 2014 at 12:42
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3$\begingroup$ $\det(I-A)=1-\mathrm{Tr}(A)+\mathrm{Tr}(\wedge^2A)+\ldots $ - but I agree this is not a question at research level. $\endgroup$– abxCommented Jan 11, 2014 at 12:59
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$\begingroup$ @Taladris: I wonder how one can know this is not a research level question. There are examples of complicated determinants which count interesting quantities such as plane partitions with particular symmetries which can be expressed as $Det(A\pm I)$ where $A$ does not seem complicated, but where finding the characteristic polynomial of $A$ does not seem easy. $\endgroup$– Douglas ZareCommented Jan 11, 2014 at 13:00
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1 Answer
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$$ \det(A+ tI) = \sum_{k=0}^n c_k^n(A).t^{n-k}, \qquad c^n_k(A) = \text{Trace}\Big(\bigwedge^k A: \bigwedge^k\mathbb R^n\to \bigwedge^k\mathbb R^n\Big). $$
answered Jan 11, 2014 at 13:00
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1$\begingroup$ @Bjørn Kjos-Hanssen: It's an exterior power, more commonly denoted as abx did in the comments with ${\bigwedge}^k$. The overset $k$ in display mode is for an AND operator. Anyway, a linear map on a vectorspace induces a multilinear map on the space of alternating $k$-forms on that vectorspace, a component of the exterior algebra. You can coordinatize it using the dimension-$k$ minors of a matrix. $\endgroup$ Commented Jan 11, 2014 at 20:57