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Suppose I have a matrix function $A(t)$ with $$\lVert A(t) - B\rVert \le ct^\alpha$$ in some matrix norm (this will work for any norm, I guess). So, in a sense $A(t)\rightarrow B$ for $t\rightarrow 0$ in $\mathcal{O}(t^\alpha)$. Plus, we have $A(0) = B$.

I happen to know the eigenvalues of $B$, but I don't know a thing about the eigenvalues of $A(t)$. Plus, $A(t)$ does not have any favorable structure, in particular, no symmetry.

So, what can you say about the eigenvalues of $A(t)$? In particular:

  • What about the spectral radius of $\lambda_{A(t)}$? Does it converge to the spectral radius of $B$?
  • Do we have $\lambda_{A(t)}\rightarrow\lambda_B$ for $t\rightarrow 0$ for all eigenvalues $\lambda_{A(t)}$ of $A(t)$?
  • And finally: is the speed of convergence $\mathcal{O}(t^\alpha)$ the same for the eigenvalues/spectral radius as for the matrix function?

The last question is actually the most important one. If the eigenvalues of $B$ are all zero, the eigenvalues as well as the spectral radius of $A(t)$ would go to zero as $\mathcal{O}(t^\alpha)$...

Any help would be appreciated, incl. references to (standard?) textbooks or papers on this matter. Maybe there is a counterexample? So far, in all numerical examples I have seen/done, all properties above do hold.

Edit: To provide a bit more background: The matrices $A(t)$ are iteration matrices which depend on a time-step size $t$. They are not this ugly, but showing convergence of this iteration has proven to be rather difficult. In the simplest case, they look like $$A(t) = (I-tQ_1)^{-1}(t(Q_2-Q_1)+B)$$ with identity matrix $I$ and some matrices $Q_1,Q_2$, which do not have any particular structure we were able to exploit so far. Now, if I can make that conclusion about the spectral radius as described above, I can state that the spectral radius is smaller than 1, i.e. the iteration converges, if the time-step size $t$ is small enough.

Edit: Does this answer help? Also, this question might be related to perturbation theory for eigenvalue problems (with non-symmetric matrices, though, and $B$ is not diagonalizable).

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Robert Speck
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3 Answers3

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Parts (1) and (2) -- Yes. The coefficients of the characteristic polynomial are continuous functions of $A(t)$ (they are polynomials in the entries of $A(t)$!) and the roots of a polynomial are continuous functions of the coefficients.

Part (3): If $B$ has non-trivial Jordan blocks, this can fail. For example, $$\begin{pmatrix} 0 & 1 \\ -t & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + O(t)$$ as $t \to 0$, but the eigenvalues are $\pm \sqrt{t}$, which is not $O(t)$.

I believe that this is true when $B$ is diagonalizable. (I know that it is diagonalizable if $B$ has distinct eigenvalues, and I'll write up that argument if this one fails.)

We may as well assume $B$ is diagonalized. Let's concentrate on a particular eigenvalue $\lambda$ of $B$, with multiplicity $m$; we might as well assume $\lambda =0$. So $B = \mathrm{diag}(0,0,\cdots,0, \lambda_2, \lambda_3, ..., \lambda_k)$ with $m$ zeroes.

Suppose that $B-A(t)$ is $O(t^a)$. Explicitly expanding the determinant, the coefficient of $x^k$ in $\det(x \mathrm{Id} - B)$ is $O(t^{a(m-k)})$ for $k < m$, and the coefficient of $x^m$ does not go to $0$ as $t \to 0$. So the Newton polygon of the characteristic polynomial passes through $(m,0)$ and stays above the line from $(m,0)$ to $(0,ma)$. This shows that the bottom $m$ roots of the characteristic polynomial vanish at rate $O(t^a)$ or faster as $t \to 0$ (and the other roots do not vanish.)

It occurs to me that it is worth sketching the argument for the Newton polygon claim directly so you can see how straightforward it is without learning the whole Newton polygon technology. Here is what I am claiming:

Lemma Fix $C>0$. Then there is a $D>0$ such that, if $f(u,z) = \sum_i f_i(u) z^i$ is holomorphic in $z$ on the disc of radius $1$, with $$|f_i(u)/f_m(u)| < \begin{cases} C u^{m-i} & i < m \\ C & i > m \end{cases}$$ then $f(u, \ )$ has $m$ roots in the disc of radius $D u$ for all sufficiently small $u$.

Proof sketch Consider $\oint \tfrac{f'}{f} dz$ where the integral is around the circle of radius $Du$. Write $f = f_m(u) z^m (1+\mbox{other terms})$ and $f' = m f_m(u) z^{m-1} (1+\mbox{other terms})$ so the integral is $\oint m \tfrac{dz}{z} + \mbox{other terms}$. Use the above conditions to bound the other terms.

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David E Speyer
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  • Thanks a lot for your elaborate answer! Also, nice counterexample! The thing is: $B$ is not diagonalizable, in the simplest case all eigenvalues of $B$ are actually zero. More precisely, $B$ is a strictly lower triangular matrix, while $A(t)$ is block-triangular, with blocks on the diagonal and stuff below. Is there anything we can do in this case? I'd be happy to provide more information, but this would require a bit more "space".. – Robert Speck Jan 03 '17 at 08:34
  • So, it sounds like your example is exactly of the form of my counter-example. I imagine the general statement should be that if $B$ has a nilpotent Jordan block of size $k$, and $A=B+O(u)$, then the corresponding eigenvalues of $A$ are $O(u^{1/k})$. Think about $\left( \begin{smallmatrix} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ t & 0 & 0 & 0 \end{smallmatrix} \right)$ to see that we can't hope for better. – David E Speyer Jan 03 '17 at 14:53
  • Does it matter that $B$ in your example would be strictly upper triangular but my $B$ is strictly lower triangular..? I guess not, right? – Robert Speck Jan 03 '17 at 15:12
  • Take the transpose of the example in my previous comment. – David E Speyer Jan 03 '17 at 15:14
  • Darn.. you're right of course. – Robert Speck Jan 03 '17 at 15:19
  • OK, since my $k\times k$-matrix $B$ has only eigenvalues zero, the eigenvalues of $A(t)$ are $\mathcal{O}(t^{\alpha/k})$.. does this sound right? – Robert Speck Jan 03 '17 at 15:26
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This is not answer. Just trying to make observations. Note that when $t$ is very close to zero, you have $$(I-tQ_1)^{-1}=I+tQ_1+t^2Q_1^2+\dots$$ Let $C=Q_2-Q1$. Then $$A(t) =(tI+t^2Q_1+t^3Q_1^2+..)C~+~(I+tQ_1+t^2Q_1^2+\dots)B $$ Now, let's look at the matrix $(I-tQ_1)^{-1}$. Let $Q_1 = T\Lambda T^{-1}$ for some invertible matrix $T$. Thus, if $\lambda_i(Q_1)$ is the $i^{th}$ eigenvalue of $Q_1$. Then we have $$\lambda_i((I-tQ_1)^{-1})=1+\lambda_it+\lambda_i^2t^2+\dots$$ Frankly, I don't know what conclusions you can draw from these. Hope it helps you or anyone else trying to answer these in some way.

dineshdileep
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The statement that eigenvalues are continuous functions of the entries of matrices is often seen in the literature. What does it really mean?

Actually, "eigenvalue continuity" is interpreted in two different meanings: in the sense of topology and in the sense of individual functions.

In the sense of topology: eigenvalues as a whole (unordered multisets) are continuous. Let A(t) approach A entrywise (say), then the eigenvalues of A(t) approach the eigenvalues of A with suitable matching. (The mapping from the matrix space to the quotient space C^n/~ is continuous.)

In the sense of individual functions: If the entries of the matrix are continuous functions over a real interval or or all the eigenvalues are real, then there is a selection of continuous functions that constitute the eigenvalues of the matrix.

Note: Eigenvalues are always continuous in the sense of topology. However, it may be impossible to parameterize eigenvalues as continuous individual functions. For example, A=[0 1; z, 0] where z belongs to a domain that contains the origin.

These can be found in Kato's book Perturbation Theory for Linear Operators and in Bhatia's book Matrix Analysis.

Fuzhen Zhang
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