Eigenvalues of symmetric matrices

Suppose A=A(x) is a smooth (or continuous) family of symmetric n\times n (real) matrices on an open set \Omega \subset \mathbb{R}^n, is it true that locally there exists a smooth (continuous) family \lambda_1 (x), \cdots, \lambda_n(x) which are the eigenvalues of A(x)?

Answer: For the continuous case, true perhaps? YES!

Proof. (Sketch, still can’t do the smooth case)

As the eigenvalues of A are the roots of the characteristic polynomial p(t)=\det(A-t I), we only have to show the continous dependence of the roots of polynomials on their coefficients. To show this, let us work in \mathbb{C} first.  Suppose p(z)=a_0+a_1 z+\cdots a_nz^n\in \mathbb{C}[z], a_n\neq 0, be a degree n polynomial. Suppose \lambda_i are the roots of p(z) (which may not be distinct). Consider one of the roots, say, \lambda_1, and choose a circle \gamma around \lambda_1, so small that it does not encloses other roots which are distinct from \lambda_1, and does not contain any other roots of p(z) on it. For another degree n polynomial q(z) close enough to p(z) (depending on \gamma and p(z)), we must have, on \gamma,

\left|\frac{q(z)}{p(z)}-1\right|<1.

So by Rouche’s theorem, if q(z) is close enough to p(z), then there must be a root of q(z) around each root of p(z) (counting multiplicity). As the number of roots of q(z) must be exactly n, this shows that for a polynomial p(z), for any \epsilon >0, there exists \delta such that if q(z)=b_0+\cdots +b_n z^n with |b_i-a_i|<\delta, for any root \lambda of p(z), there exists a root \tilde \lambda of q(z) such that |\tilde \lambda-\lambda|<\epsilon.

Note that for any symmetric n\times n matrix B, its characteristic must have n real roots. So setting the polynomial q(z) above to be the characteristic polynomial of B (and p(z) to be the real characteristic polynomial of A), we conclude that if B is close enough to A, the eigenvalues of B are real and are also close to A.

Now, given a continuous family A(x), if we define \lambda_1(x)\leq \lambda_2(x)\leq \cdots \leq \lambda_n(x) to be the eigenvalues of A(x), by the above analysis, \lambda_i(x) would be continuous at each point x.

Remark: \lambda_i defined in this way may not be smooth, even A(x) is smooth, e.g.

A(t)=\begin{pmatrix}&t &0\\&0 &1-t\end{pmatrix} for 0<t<1.

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2 Responses to Eigenvalues of symmetric matrices

  1. kingleunglee says:

    What is the meaning of smooth? For example, what is A'(x) . Is it if A(x)=(a_{ij}(x)), then A'(x)=(a'_{ij}(x))?

  2. kkkwong says:

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

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