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A simple remark on the spectrum of certain matrix sums

2 October, 2009

Having let the Shavgulidze-Thompson project slide out of the intray and into the mountain of Unfinished Loose Ends, I feel I should compensate with something vaguely mathematical. Hence this post, which is a follow up to some comments I left on a post at the Secret Blogging Seminar.

More precisely, in response to Q2 on that post, I left some rather dim-witted and error-strewn comments, only to have light shed by this subsequent observation from Greg Kuperberg:

Proposition: Let A and B be two Hermitian matrices. Then the spectrum of A+iB lies in the rectangle formed by the first and last eigenvalues of A and B.

Once GK stated the correct result, I realised that it followed from some facts that I really should have known – or knew, but had momentarily forgotten. It seems that the argument I had in mind is slightly different, at least in presentation, from the proof GK had in mind, and so I thought I’d give it here. (His reasoning seems like it should be more robust, and extend more easily to the case of bounded operators on infinite-dimensional Hilbert space.)

Claim. Let A and B be normal matrices, with spectra \sigma(A) and \sigma(B) respectively. Then the spectrum of A+B is contained in {\rm co}\, \sigma(A)+{\rm co}\,\sigma(B).

Proof. Since A is normal, there exists an orthonormal basis of {\bf C}^n, which consists of eigenvectors for A. Let’s denote this basis by v_1,\dots,v_n and let the corresponding eigenvalues be \lambda_1,\dots,\lambda_n.

Similarly, there is an orthonormal basis w_1,\dots, w_n and scalars \mu_1,\dots,\mu_n such that Bw_k=\mu_kw_k for all j.

Now let \alpha be an eigenvalue of A+B, and let x be a corresponding eigenvector of unit length. We have

\alpha = \langle \alpha x , x\rangle = \langle (A+B)x,x\rangle = \langle Ax,x\rangle + \langle Bx, x\rangle

But now we can exploit the fact that A and B each have a complete set of orthonormal eigenvectors. In particular, writing x = \sum_j \langle x, v_j \rangle v_j we have

\langle Ax , x \rangle = \sum_j \lambda_j \vert \langle x, v_j \rangle \vert^2

We have \sum_j \vert\langle x, v_j\rangle\vert^2=1 (again, using the orthonormality of the v_j) and so \langle Ax,x\rangle \in {\rm co}\, \{ \lambda_1,\dots,\lambda_n\} = {\rm co}\sigma(A). An exactly similar argument, this time using the w_k, tells us that \langle Bx, x\rangle \in {\rm co}\,\sigma(B). Hence \alpha=\langle Ax,x\rangle+ \langle Bx,x\rangle lies in the sum of these two convex hulls, as claimed.

Cards on the table, or the man behind the curtain

I have to confess that the phrasing of the argument above wasn’t the first that came to mind when I read GK’s comment. Lurking in the background — above and, I suspect, in his approach also — is the concept of numerical range. The numerical range of an n\times n complex matrix M is the set

W(A)=\{ \langle Mx ,x \rangle | x \in {\bf C}^n, \|x\|_2 \leq 1 \}

and it is clear that W(A+B) is contained in W(A)+W(B) for every pair A, B of n\times n matrices. Now, by considering appropriate eigenvectors, one sees that every eigenvalue of M is contained in W(M). Also, if D is a diagonal matrix, then the same calculation that was made above shows that W(D) is contained in the convex hull of \sigma(D), and since the numerical range is unchanged if we conjugate by a unitary matrix, it follows that W(M)\subseteq {\rm co}\, \sigma(M) for every normal matrix M. In particular, if A and B are normal n\times n matrices then

\sigma(A+B) \subseteq W(A+B) \subseteq W(A)+W(B) = {\rm co}\,\sigma(A)+{\rm co}\,\sigma(B)

which is in effect what we proved above.

The reason I should have remembered this is that a short while back I was reading up on some aspects of the numerical range for operators on infinite-dimensional spaces. The definition is the obvious one, and what is interesting is that we still have

  1. \sigma(T)\subseteq W(T) for every bounded operator T;
  2. W(M) = {\rm co}\,\sigma(M) for every normal operator M.

Note that in infinite dimensions the spectrum of T might contain points which are not eigenvalues, and so the argument above with eigenvectors doesn’t work anymore.

2 Comments leave one →
  1. Minghua permalink
    5 March, 2010 4:38 pm

    What about the case of pectrum of parallel sum of two matrices?

    • 5 March, 2010 5:03 pm

      I’m not sure what you mean by “parallel sum”. Could you please give an example?

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