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Say you have 2 symmetric matrices, $A$ and $B$, and you know that every linear combination $xA+yB$ ($x,\\,y\in \mathbb{R}$) has an eigenvalue of multiplicity at least $m>1$. Such a situation can of course be obtained if $A$, $B$ have a common eigenspace of multiplicity at least $m$.

My question is: is it the only possibility?

A way to proceed is the following: the characteristic polynomia of the generic matrix is $\det(xA+yB-tI)$, and its discriminant $\Delta$ (with respect to $t$) is a homogeneous polynomial in $x,y$ of degree $n^2-n$, where $n$ is the number of rows and columns in $A$ and $B$. Since every matrix in the family has some eigenvalue of multiplicity $>1$, the polynomial $\Delta$ vanishes identically, hance all the $n^2-n+1$ coefficients in $\Delta$ vanish. This gives $n^2-n+1$ polynomial conditions on the $n^2+n$ coefficients of $A$ and $B$, and this might help somehow.

Still, both finding these polynomial conditions and solving them, seems to be painful and extremely computational. Maybe there are better ways to proceed $\ldots$?

Thanks in advance!

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3 Answers 3

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This is false. Let $A_0$ and $B_0$ be $k \times k$ symmetric matrices with no common eigenspace. Let $A$ and $B$ be the $2k \times 2k$ matrices with block forms $\left( \begin{smallmatrix} A_0 & 0 \\ 0 & A_0 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} B_0 & 0 \\ 0 & B_0 \end{smallmatrix} \right)$. Then $A$ and $B$ have no common eigenspace, but every eigenvalue of $Ax+By$ has multiplicity $\geq 2$. And, of course, you can replace $2$ by any $m$.

Here is a different example. Let $A$ and $B$ be symmetric matrices of the form $$\begin{pmatrix} \ast & \ast & \ast & \ast & \ast & \ast \\ \ast & \ast & \ast & \ast & \ast & \ast \\ \ast & \ast & 0 & 0 & 0 & 0 \\ \ast & \ast & 0 & 0 & 0 & 0 \\ \ast & \ast & 0 & 0 & 0 & 0 \\ \ast & \ast & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

Then $Ax+By$ is also of this form, and you can check that any matrix of this form has a zero eigenspace of multiplicity $\geq 2$. However, there is no reason for the $0$-eigenspaces of $A$ and $B$ to meet at all!


More generally, regarding "a better way": For any matrices $A$ and $B$, let $F(x,y,z) = \det(z \mathrm{Id} + x A + y B)$. This defines a curve in $\mathbb{P}^2$. Your condition is every line through $(0:0:1)$ intersects this curve in a point of multiplicity $\geq m$. This turns out to imply that $F$ must have a component of multiplicity $\geq m$: i.e. $F$ factors as $G^m H$ for some homogenous polynomials $G$ and $H$.

If $A$ and $B$ had a common eigenspace, then $G$ would be linear, and $A$ and $B$ would have direct sum decompositions $A = A_1^{\oplus m} \oplus A_2$ and $B = B_1^{\oplus m} \oplus B_2$ where $(A_1, B_1)$ and $(A_2, B_2)$ give rise to the plane curves $G$ and $H$.

The preceeding examples show that life can be more complicated. In the first example, $F=G^2$, for some nonlinear $G$. In the second, the factorization of $F$ does not correspond to a direct sum decomposition of $(A,B)$.

There is a fair amount of literature on expressing plane curves as determinants, so there is probably someone who has looked specifically in the case of a multiple factor, but I don't know who.

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It makes sense! thanks! –  Marco Radeschi Jun 30 '11 at 20:09
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If you restrict to the postulated $m$-dimensional eigenspace $E$ of $C=xA + yB,$ the fact that $C+\epsilon A$ has the same multiplicity, means that $A$ restriced to $E$ is also a multiple of identity, as is $B$ (by the same argument). Since a small perturbation of a matrix with distinct eigenvalues still has distinct eigenvalues, the situation in the previous sentence must occur for your condition to hold.

I strongly suggest you read the first couple of chapters of Kato's "Perturbation theory of linear operators" (the first two chapters deal with the finite-dimensional case).

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I'm sorry, I didn't quite get your answer. What do you mean by "common eigenspace"? I'm not assuming any common eigenspace, how can you prove that it exists? –  Marco Radeschi Jun 30 '11 at 17:22
    
I misspoke: I meant the $m$-dimensional eigenspace, will correct. –  Igor Rivin Jun 30 '11 at 17:52
    
One more question: from your answer it sound like you know that there is a fixed eigenspace $E$, such that $C+\epsilon A$ sends $E$ to itself (at least for $\epsilon$ small enough i guess...) Why is this true? Assumed this, I understand your argument, but i don't know why this is true.. –  Marco Radeschi Jun 30 '11 at 18:29
    
@Marco: the point is that locally the multiplicity of an eigenvalue cannot increase, it can only decrease, so the only way the multiplicity can stay at $m$ for a very small $\epsilon$ is when $C$ and $A$ and $B$ all commute when restricted to $E.$ Do look at Kato (@Chris is right, the first two chapters of Kato are not that short, but are quite well-written). –  Igor Rivin Jun 30 '11 at 18:56
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I understand that point, but i don't understand why in principle the eigenspace couldn't vary with respect to $\epsilon$... using Kato's notation, this should amount to saying that the eigenprojection $P_{\epsilon}$ is constant... but maybe i still don't know some facts, i am reading at Kato's book right now (thanks for the reference!) –  Marco Radeschi Jun 30 '11 at 19:22
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I think that the most interesting examples come from systems of conservation laws in physics. These are systems of PDEs (partial differential equations). When they are first-order and linear (linearity can be achived by linearizing about a constant state), you have a system $$\partial_tU+\sum_{\alpha=1}^dA^\alpha U=f,$$ where $A^\alpha\in{\bf M}_n({\mathbb R})$. The space dimension is usually $d=3$ but may be smaller for waves propagating in specific directions. Exemples cover the Euler equations for gas dynamics, the Maxwell's equation for electro-magnetism, Magneto-hydrodynamics, elasticity and so on.

The symbol $A(\xi)=\sum_\alpha A^\alpha$ plays a fundamental role. Its eigenvalues are the wave velocities, times the modulus $|\xi|$. In several exemples, especially when the system has a group invariance, the eigenvalues have constant multiplicities, yet the eigenspace do vary with $\xi$. Let us take the example of Maxwell's equations in vacuum. Here $d=3$ and $n=6$. We have $$A(\xi)=\begin{pmatrix} 0_3 & J(\xi) \\\\ -J(\xi) & 0_3 \end{pmatrix},$$ where $J(\xi)$ is the matrix of the cross product by $\xi$: $J(\xi)X=\xi\times X$. The eigenvalues of $A(\xi)$ are $0$ and $\pm|\xi|$. None of the eigenvectors is constant. In particular, two matrices $A^\alpha$ don't have a common eigenspace.

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