most recent 30 from http://mathoverflow.net 2013-05-25T11:29:45Z http://mathoverflow.net/feeds/question/8331 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8331/approximately-known-matrix Daniel Moskovich 2009-12-09T08:18:30Z 2009-12-13T14:13:42Z

<p>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).</p>

http://mathoverflow.net/questions/8331/approximately-known-matrix/8335#8335 Thorny 2009-12-09T10:04:04Z 2009-12-09T10:04:04Z

<p>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.</p>

http://mathoverflow.net/questions/8331/approximately-known-matrix/8336#8336 David Lehavi 2009-12-09T10:04:27Z 2009-12-09T15:03:14Z

<p>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).</p>

http://mathoverflow.net/questions/8331/approximately-known-matrix/8755#8755 noob 2009-12-13T14:13:42Z 2009-12-13T14:13:42Z

<p>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.</p>