-2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ What are you expecting? A magic formula where none exists? $\endgroup$ Jan 11, 2014 at 12:36
  • 1
    $\begingroup$ @MichaelRenardy: I wonder how one can know that a formula does not exist. Anyway, the OP is not a research level question. $\endgroup$
    – Taladris
    Jan 11, 2014 at 12:42
  • 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. $\endgroup$
    – abx
    Jan 11, 2014 at 12:59
  • $\begingroup$ @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. $\endgroup$ Jan 11, 2014 at 13:00

1 Answer 1

2
$\begingroup$

$$ \det(A+ tI) = \sum_{k=0}^n c_k^n(A).t^{n-k}, \qquad c^n_k(A) = \text{Trace}\Big(\bigwedge^k A: \bigwedge^k\mathbb R^n\to \bigwedge^k\mathbb R^n\Big). $$

$\endgroup$
2
  • $\begingroup$ What's $\wedge $? $\endgroup$ Jan 11, 2014 at 16:56
  • 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. $\endgroup$ Jan 11, 2014 at 20:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.