## Approximately known matrix

What linear algebraic quantities can be calculated precisely for a nonsingular matrix whose entries are only approximately known (say, entries in the matrix are all huge numbers, known up to an accuracy of plus or minus some small number)? Clearly not the determinant or the trace, but probably the signature, and maybe some sort of twisted signatures? What is a reference for this sort of stuff? (numerical linear algebra, my guess for the name of such a field, seems to mean something else).

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SVD is stable, and in some sense incorporates all the stable data you can have, so the answer is: "anything you can see on the SVD". Specifically you can easily see the signature (assuming the matrix is far enough from being singular).

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 Could you add some details? What do you mean by "SVD is stable"? – Daniel Moskovich Dec 9 2009 at 10:11 I mean that you can easily bound the change in the output by the change in the input (and the input itself); which is something you cannot do for e.g. LU decomposition. Caveat: If some Eigenvalue of the SVD is (close to) multiple, then the output is of course the span of the related vectors, and not the vectors themselves. – David Lehavi Dec 9 2009 at 11:05 I assume you mean “far enough from _singular_”. – Harald Hanche-Olsen Dec 9 2009 at 13:59 @ Harald: Thanks – David Lehavi Dec 9 2009 at 15:02 This is a great answer. What about the change in the singular values? Also, is there a name for this field or a reference for related problems? – Daniel Moskovich Dec 10 2009 at 4:24
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If an invariant of nonsingular matrices is locally constant (I guess this is what's meant by "can be calculated precisely"), then it can only depend on the connected component of the linear group, which means only the orientation (sign of the determinant) can be calculated. For symmetric matrices, the same argument shows that any calculable quantity is a function of the signature since any matrix can be connected to a standard representatives of one of the signature classes using a continuous version of orthogonalization.

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I think the words "approximate" and "numerical" in the question hint that this is not really what Daniel has in mind.... – David Lehavi Dec 9 2009 at 15:15
This is a fair interpretation of "precisely," I think. Perhaps Daniel should rephrase his question if that's not what he has in mind. – Qiaochu Yuan Dec 9 2009 at 15:50

The name "matrix analysis" seems to be associated with questions like this. This is an answer instead of a comment because I lack brownie points.

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