# Eigenvalues of matrix with entries that are continuous functions

For each $t \in [0,b]$, let $M(t)$ be an $n \times n$ matrix with entries $m_{ij}(t).$ The matrix $M(t)$ is invertible and positive-definite, so the eigenvalues of $M(t)$ exist and are positive for every $t$.

The entries are continuous functions: $m_{ij} \in C^0[0,b]$. Does this mean the eigenvalues $\lambda_i(t)$ of $M(t)$ are also continuous functions?

It is true for $n=1$ and $n=2$. I can't do it for the general case.

• Linear operators do have eigenvalues, so from which vector space is your Matrix ? – Dominic Michaelis Aug 31 '13 at 13:25
• $M(t)$ is just an $n \times n$ matrix for each $t \in [0,b].$ So just $\mathbb{R}^{n \times n}.$ I just want to know if the eigenvalues are continuous wrt. $t$. – ampeo Aug 31 '13 at 13:30
• a matrix does have many eigenvalues. in which sense you mean continuous ? – Dominic Michaelis Aug 31 '13 at 13:36
• I guess that you mean the following : $M(t_0)$ is given, are the eigenvalues of $M(t_0+h)$ close to the one of $M(t_0)$ for small values of $h$ in the sense of the metric ? (edit : and the second question is what do you want to prove at the end of the day exactly ? (in term of mathematical proposition)) – Bertrand R Aug 31 '13 at 13:45
• @BertrandR thats what i thought first but what to do if the matrix doesn't have any eigenvalues ? – Dominic Michaelis Aug 31 '13 at 13:49

The eigenvalues of $M(t)$ are continuous functions of $t$, and this holds whether or not $M(t)$ is invertible and/or positive definite.

It follows from the fact that the roots of any polynomial $p(z) \in \Bbb C[z]$ are continuous functions of the coefficients $p_i \in \Bbb C$, where we have

$p(z) =\sum_{i = 0}^{i = \deg p} p_i z^i$.

This result is in fact quite well known. An excellent paper covering it and more, similar results, is Continuity and Location of Zeroes of Linear Combinations of Polynomials, by Mishael Zedek, Proc. Amer. Math. Soc. 16 (1965), 78-84, a copy of which is available for free download here.

Now consider the "characteristic polynomial" $p(\mu, t) = \det (\mu I - M(t))$; this is in fact a monic polynomial in $C^0[0, b][\mu]$, the ring of polynomials with coefficients in $C[0, b]$; indeed, for each individual $t \in [0, b]$, $p(\mu, t)$ is in fact the usual characteristic polynomial of $M(t)$. To actually see that $p(\mu, t) \in C^0[0, b][\mu]$, simply note that the coefficients $p_i(t)$ of $p(\mu, t) = \sum_0^n p_i(t) \mu^i$ depend continuously on the $m_{ij}(t)$, being themselves in fact mulitvariate polynomials in the coefficients $m_{ij}(t)$ of the matrix $M(t)$; and since the $m_{ij}(t)$ are themselves continuous, we simply invoke that old adage ($\equiv$, in the present context, well-known mathematical truth) that a continuous function of a continuous function is continuous to conclude that we have $p_i(t) \in C^0[0, b]$ for all $i$, $0 \le i \le n$. We see that in fact the the one-parameter family of characteristic polynomials $p(\mu, t) = \det (\mu I - M(t))$ has coefficients continuous in $t$; thus the zereos $\mu(t)$ of $p(\mu,t)$, i.e. the eigenvalues of $M(t) = [m_{ij}(t)]$ are themselves continuous functions of $t$ by the result of Zedek cited above.

In closing, I should add the following note: in the event that the $m_{ij}(t) \in C^1[0. b]$, i.e., are differentiable functions of $t$, we can say a lot more; for then, the coefficients $p_i(t)$ of the characteristic polynomial will themselves be differentiable functions of $t$; we can thus use the implicit function theorem applied to the equation $p(\mu, t) = \det(\mu I - M(t))$ to conclude that, as long as

$\frac {\partial p}{\partial \mu}(\mu, t) = \sum_1^n p_i(t)i \mu^{i - 1} \ne 0 \tag{A}$

the eigenvalue functions $\mu(t)$ are in fact differentiable. We can even compute their derivative: since

$p(\mu(t), t) = \sum_0^n p_i(t) \mu^i(t) = 0$,

taking the $t$-derivative (denoted by the "dot") yields

$\dot p(\mu, t) + \frac{\partial p}{\partial \mu} \dot{\mu}(t) = \sum_0^n \dot{p}_i(t) \mu^i(t) + (\sum_1^n p_i(t)i \mu^{i - 1}) \dot{\mu}(t) =0; \tag{B}$

when (A) holds, (B) can obviously be solved for $\dot \mu(t)$; and (A) is precisely the condition that $\mu(t)$ is only a simple zero of $p(\mu(t), t)$, that is, of multiplicity one. In this case we can actually track the motion of $\mu(t)$ by looking at $\dot \mu(t)$, which will work until the multiplicity of $\mu(t)$ jumps up to two or more; this happens when the zeroes of $p(\mu, t)$ "collide".

Fun stuff, and useful, too, I'd type all afternoon but my night job beckons, gotta run!

Hope this helps!

Cheers!

• Thank you, a great answer! – ampeo Sep 1 '13 at 8:16
• @RobertLewis I know this is an old thread but I'll give it a shot. I was wondering if you have a reference for the second part of your beautiful answer. That is, a reference I could use to show that if the entries of a matrix are differentiable with respect to some parameters, then the eigenvalue functions are also differentiable. Thanks in advance. – rpa Apr 7 at 21:58
• @rpa: for example, I just googled with the search phrase, "differentiable dependence of eigenvalues on parameters" and got some good-looking leads. I'm sure one could do even better by refining the search. There seem to be quite a few journal articles out there on this! – Robert Lewis Apr 7 at 22:11
• @RobertLewis Thank you! I will look around. Please let me know if you remember where you saw it. I would really appreciate it. Have a great day! – rpa Apr 7 at 22:11
• @RobertLewis. I found the statement I was looking for in p. 116 of Pertubation Theory for Linear Operators by Tosio Kato. Thanks a lot for your help! – rpa Apr 7 at 22:28