Suppose is a smooth (or continuous) family of symmetric (real) matrices on an open set , is it true that locally there exists a smooth (continuous) family which are the eigenvalues of ?
Answer: For the continuous case, true perhaps? YES!
Proof. (Sketch, still can’t do the smooth case)
As the eigenvalues of are the roots of the characteristic polynomial , we only have to show the continous dependence of the roots of polynomials on their coefficients. To show this, let us work in first. Suppose , , be a degree n polynomial. Suppose are the roots of (which may not be distinct). Consider one of the roots, say, , and choose a circle around , so small that it does not encloses other roots which are distinct from , and does not contain any other roots of on it. For another degree polynomial close enough to (depending on and ), we must have, on ,
So by Rouche’s theorem, if is close enough to , then there must be a root of around each root of (counting multiplicity). As the number of roots of must be exactly , this shows that for a polynomial , for any , there exists such that if with , for any root of , there exists a root of such that .
Note that for any symmetric matrix , its characteristic must have real roots. So setting the polynomial above to be the characteristic polynomial of (and to be the real characteristic polynomial of A), we conclude that if is close enough to , the eigenvalues of are real and are also close to .
Now, given a continuous family , if we define to be the eigenvalues of , by the above analysis, would be continuous at each point .
Remark: defined in this way may not be smooth, even is smooth, e.g.