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,


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.

This entry was posted in Algebra, Linear Algebra. Bookmark the permalink.

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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