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.

for .

What is the meaning of smooth? For example, what is . Is it if , then ?

It means its entries are smooth functions of , as you just wrote.