I have a set of $N\times N$ hermitian matrices $A_i,~i=1,\dots,M$. Are there any results on the possibility of simultaneously tridiagonalizing them?

$\begingroup$ for $M=2$, see sciencedirect.com/science/article/pii/S0021869308000045 $\endgroup$ – Carlo Beenakker Jun 15 '14 at 6:29

$\begingroup$ thanks for the reply. But I was interested in the general case. $\endgroup$ – dineshdileep Jun 19 '14 at 8:06
Not an answer, but rather a pointer to some potentially interesting references. Two relevant geometric ideas are flag varieties (or manifolds) and Hessenberg varieties.
Flags connect abstract linear operators with matrices, with possible various structures. A matrix differs from an abstract linear operator by being expressed in a particular basis, say $e_i$. Often, it is not the basis itself that is important, but the subspaces $V_i = \operatorname{span}(e_j)_{j=1}^i$. In general, a sequence of subspaces $0 \subset V_1 \subset V_2 \subset \cdots \subset V_{n1} \subset \mathbb{R}^n$, with the dimension of each $V_i$ increasing by one is called a complete flag. Picking a vector $e_i \in V_i \setminus V_{i1}$ generates, in the reverse direction, a basis from a flag. It so happens that the space of all complete flags $\mathrm{Flag}_n$ is a smooth compact manifold (even an algebraic variety) of dimension $n(n1)/2$.
Given a linear operator $A\colon \mathbb{R}^n \to \mathbb{R}^n$, we can consider special flags that satisfy $A V_i \subseteq V_{i+1}$. These are called Hessenberg flags, because the matrix corresponding to $A$ in a basis adapted to the flag is in upper Hessenberg form: all entries below the submain diagonal are zero ($a_{ij} = 0$ if $i > j+1$). If $A$ happens to be symmetric ($a_{ij} = a_{ji}$), then being in upper Hessenberg form is the same as being in tridiagonal form. The set of all Hessenberg flags of $A$ is then a subset $\mathrm{Hess}(A) \subset \mathrm{Flag}_n$, referred to as the Hessenberg variety of $A$. It seems that the study of the geometry and topology of Hessenberg varieties was initiated in
 On the topology of the Hessenberg varieties of a matrix Filippo de Mari Casareto dal Verme, PhD Dissertation Washington University 75 (1987)
 Generalized Eulerian numbers and the topology of the Hessenberg variety of a matrix, Filippo de Mari, Mark A. Shayman, Acta Applicandae Mathematica (1988) 12 213235
According to these references, $\dim \mathrm{Hess}(A) = n1$ for a generic operator $A$. Since $2(n1) < n(n1)/2$ for $n > 4$, we cannot expect $\mathrm{Hess}(A)$ and $\mathrm{Hess}(B)$, for generic $A$ and $B$, to have nonempty intersection in $\mathrm{Flag}_n$ for $n > 4$. So in higher dimensions. One cannot exect any two random operators to be simultaneously reducible to upper Hessenberg form. Of course, there may exist some simple condition on $A$ and $B$ such that guarantees simultaneous reducibility to upper Hessenberg form, but I haven't thought of it.
As already noted, for symmetric operators, upper Hessenberg and tridiagonal forms are related. So, if one were to think of a set of conditions for simultaneous upper Hessenberg form, they might lead to similar conditions for simultaneous tridiagonalizability. But, in the absence of such conditions I'm afraid I'm no closer to answering your question.

$\begingroup$ are you suggesting that for $n\leq 5$, simultaneous tridiagonalization is possible? $\endgroup$ – dineshdileep Jul 13 '14 at 2:14

$\begingroup$ It seems so, just by dimension counting. To submanifolds whose dimensions sum to more than the dimension of the ambient space generically intersect each other. $\endgroup$ – Igor Khavkine Jul 13 '14 at 17:54


$\begingroup$ Sorry, had an offbyone error. It should be $n\le 4$ not $5$ Again, counting the dimensions, you have $M_2 = \infty$, $M_3 = 3$ and $M_4 = 2$, where $M_n$ is the maximum number of $n\times n$ matrices you can expect to simultaneously put in upper Hessenberg form (hence tridiagonal for, if they are symmetric). $\endgroup$ – Igor Khavkine Jul 14 '14 at 9:23