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!



Multiplicity of eigenvalues in 2-dim families of symmetric matrices
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!




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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.


