Multiplicity of eigenvalues in 2-dim families of symmetric matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:25:25Z http://mathoverflow.net/feeds/question/69187 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69187/multiplicity-of-eigenvalues-in-2-dim-families-of-symmetric-matrices Multiplicity of eigenvalues in 2-dim families of symmetric matrices Marco Radeschi 2011-06-30T14:22:33Z 2011-07-01T06:22:49Z <p>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 $m$.</p> <p>My question is: is it the only possibility?</p> <p>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.</p> <p>Still, both finding these polynomial conditions and solving them, seems to be painful and extremely computational. Maybe there are better ways to proceed $\ldots$?</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/69187/multiplicity-of-eigenvalues-in-2-dim-families-of-symmetric-matrices/69204#69204 Answer by Igor Rivin for Multiplicity of eigenvalues in 2-dim families of symmetric matrices Igor Rivin 2011-06-30T17:14:57Z 2011-06-30T17:53:13Z <p>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.</p> <p>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). </p> http://mathoverflow.net/questions/69187/multiplicity-of-eigenvalues-in-2-dim-families-of-symmetric-matrices/69214#69214 Answer by David Speyer for Multiplicity of eigenvalues in 2-dim families of symmetric matrices David Speyer 2011-06-30T20:05:57Z 2011-06-30T20:34:02Z <p>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 <code>$\left( \begin{smallmatrix} A_0 &amp; 0 \\ 0 &amp; A_0 \end{smallmatrix} \right)$</code> and <code>$\left( \begin{smallmatrix} B_0 &amp; 0 \\ 0 &amp; B_0 \end{smallmatrix} \right)$</code>. 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 $m$.</p> <p>Here is a different example. Let $A$ and $B$ be symmetric matrices of the form <code>$$\begin{pmatrix} \ast &amp; \ast &amp; \ast &amp; \ast &amp; \ast &amp; \ast \\ \ast &amp; \ast &amp; \ast &amp; \ast &amp; \ast &amp; \ast \\ \ast &amp; \ast &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ \ast &amp; \ast &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ \ast &amp; \ast &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ \ast &amp; \ast &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ \end{pmatrix}$$</code></p> <p>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!</p> <hr> <p>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 $H$. </p> <p>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 $H$.</p> <p>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 $(A,B)$. </p> <p>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.</p> http://mathoverflow.net/questions/69187/multiplicity-of-eigenvalues-in-2-dim-families-of-symmetric-matrices/69234#69234 Answer by Denis Serre for Multiplicity of eigenvalues in 2-dim families of symmetric matrices Denis Serre 2011-07-01T06:22:49Z 2011-07-01T06:22:49Z <p>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.</p> <p>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 $$A(\xi)=\begin{pmatrix} 0_3 &amp; J(\xi) \\ -J(\xi) &amp; 0_3 \end{pmatrix},$$ 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.</p>