Sturm sequences

The matrix elements of each transfer matrix are polynomials with respect to the eigenvalue $\lambda$, beginning with the basic matrix whose 11 element is of the first degree in $\lambda$, which is absent from the remaining elements. Multiplying transfer matrices raises the degree of the matrix elements, the 11 element always retaining the highest degree - the sum of the degrees of the factors - when the matrices have the form we are using. Boundary conditions eventually dictate that some combination of matrix elements gives zero, which means the vanishment of a polynomial whose degree is the length of the string, whose expression is evidently the characteristic equation of $A$.

If the boundary condition is independent of the length of the string, there is a series of characteristic equations, each having a similar structure yet differing by the degree of the polynomial involved. Moreover, the pecuiliar structure of the matrix $M$, arising from the tautology which was incorporated in its definition, interlaces the polynomials used at each stage. It is worth seeing how this is done, in the simple case where the boundary condition is that there be a terminal node.

$$\begin{eqnarray*} M_{i+1} & = & \left[ \begin{array}{cc} a \lambda - b & - ... ...mbda - b) P_{21}^i \\ P_{11}^i & P_{12}^i \end{array} \right] \end{eqnarray*}$$

If the initial vector were $(1 0)$, implying $x_1 = 0, x_2 = 1$ (any factor will do), and the final vector were $(0, 1)$ implying that $x_{i+1}$ were also zero, then the 11 element of $M_{i+1}$ would vanish, or

$$\begin{eqnarray*} (a \lambda - b) P_{11}^i - c P_{21}^i & = & 0, \\ a \lambda & = & b + c \frac{P_{21}^i }{ P_{11}^i } \end{eqnarray*}$$

Figure: Interleaving of eigenvalues

The new roots lie on a line determined by the diagonal element of $A$ with slope likewise determined by $A$ - the ratio of its diagonal to off-diagonal elements. The roots also lie on the quotient of two polynomials; the zeroes of the denominator are just the roots of the next smaller $A$, the roots of the numerator are those of the second previous $A$. Not both polynomials have roots in the same place because of the limit on the maximal thickness of a node. Thus zeros and poles of the quotient alternate, and the new roots lie between the poles so they are also separated. Thus all roots are distinct, and interleaved; this is called a Sturm sequence.

There are interesting details in the graph which has been sketched. The derivative of the quotient is nonzero and always has a consistent sign, making it monotone within the panels delimited by its poles. Consequently the intersecting line crosses the quotient just once in each panel, assuring the interleaving of roots and keeping them real. The root cause of all this is the sign-symmetry of the matrix $A$; were it sign-antisymmetric, roots would have to run along in conjugate pairs. In fact, the interleaving property was already proved on more general grounds by Ledermann, so that we mainly have a more graphic presentation of the result.

The value where the line crosses the axis is the diagonal element of the bordering matrix, which will not be an eigenvalue although it will necessarily sit between some pair of old roots. It acts as a center of readjustment, crowding new eigenvalues toward the edges of the panels defined by the old eigenvalues while distancing them from still older eigenvalues. Eventually there will be very little change in eigenvalues far from that diagonal element since the line will intersect the quotient near its asymptotes, but finally one of the new eigenvalues will lie outside the range established previously.