# Similarity of two tridiagonal matrices

I am considering two complex symmetric tridiagonal matrices. First, A is a tridiagonal matrix with identical non-diagonal elements :

A = $$\begin{pmatrix} ig_1 & \kappa & 0 & 0 \\ \kappa & -ig_2 & \kappa & 0 \\ 0 & \kappa & -ig_1 & \kappa \\ 0 & 0 & \kappa & i g_2 \end{pmatrix}$$

A second matrix B has non-identical non-diagonal elements and is a function of some control parameters $$X = [x_1, x_2, x_3, x_4]^T$$:

B(X) = $$\begin{pmatrix} ix_1 & \kappa_1 & 0 & 0 \\ \kappa_1 & ix_2 & \kappa_2 & 0 \\ 0 & \kappa_2 & ix_3 & \kappa_3 \\ 0 & 0 & \kappa_3 & ix_4 \end{pmatrix}$$

The problem I am trying to solve is finding X such that $$B(X) \sim A$$.

So far, I have look into Jordan decomposition. I used $$T_A J_A = A T_A$$ and $$T_{B^T} J_B = B^T T_{B^T}$$ (in which T and J correspond to the transformation and Jordan matrices). If I note $$S = T_A {T_{B^T}^T}$$, the problem reduces to:

$$X \, | \, SB(X) - AS = 0$$

, which I can put into a numerical code to solve.

What I am looking for would be another solution, which I could use to derive an analytical formula to find X as a function of the different parameters. Does anyone has something to suggest? Maybe a different type of decomposition, like LU?

Best

As they are 3-diagonal, it's a shame to lose this property and attack it with full formalism (like LU). Complex symmetric matrices are not guaranteed to even be diagonalizable, but you can still require the characteristic polynomials of both matrices to be equal as a first necessary condition. Before even computing the polynomials, you can require the trace and determinant of both matrices to match:

$$\operatorname{Tr} A=0$$ $$\det A = g_1^2g_2^2-g_1g_2 \kappa^2+\kappa^4$$

And finally, the full polynomial; for a tridiagonal matrix, in the definition of the determinant the only permutations permitted are those with elements displaced at most by one, so there are not a lot of options, just (1234), (1243), (1324), (2134), (2143). Additionally, the symmetries of the first matrix make it assume a very special form (which also includes the quick conditions above):

$$\det (A-\lambda I)=(\lambda^2-g_1^2)(\lambda^2-g_2^2) -\kappa^2(3\lambda^2-i (g_1+g_2)\lambda+g_1g_2)+\kappa^4=0$$

On the second matrix, the equality of characteristic polynomial will give you 4 equations to fulfill, so unless some of them happen to be trivially true, you have 4 conditions for 4 parameters which should determine the system exactly, or prove it is impossible to do. You immediately see the linear equation from trace condition (or absence of $$\lambda^3$$ term), and the rest follow:

$$x_1+x_2+x_3+x_4=0$$

$$x_1x_2x_3x_4+\kappa_1^2 x_3 x_4+\kappa_2^2 x_1 x_4+\kappa_3^2 x_1 x_2+\kappa_1^2\kappa_3^2=g_1^2g_2^2-g_1g_2 \kappa^2+\kappa^4$$

$$x_1x_2x_3+x_1 x_2 x_4+x_1 x_3 x_4+x_2 x_3 x_4+\kappa_1^2(x_3+x_4)+\kappa_2^2 (x_1+x_4)+\kappa_3^2(x_1+x_2)=\kappa^2(g_1+g_2)$$

$$x_1x_2+x_1 x_3+x_2 x_3+x_1 x_4+x_2 x_4+x_3 x_4+\kappa_1^2+\kappa_2^2+\kappa_3^2=3\kappa^2-g_1^2-g_2^2$$

The first condition is the most helpful, because squaring it simplifies the last condition:

$$x_1x_2+x_1 x_3+x_2 x_3+x_1 x_4+x_2 x_4+x_3 x_4=-\frac{1}{2}(x_1^2+x_2^2+x_3^2+x_4^2)=-\frac{||\vec{x}||^2}{2}$$

$$-\frac{||\vec{x}||^2}{2}=3\kappa^2-\kappa_1^2-\kappa_2^2-\kappa_3^2-g_1^2-g_2^2$$

Similar manipulation of cubic term gives you a similar simplification: $$x_1x_2x_3+x_1 x_2 x_4+x_1 x_3 x_4+x_2 x_3 x_4=\frac{1}{3}(x_1^3+x_2^3+x_3^3+x_4^3)$$

$$\frac{1}{3}(x_1^3+x_2^3+x_3^3+x_4^3)+\kappa_1^2(x_3+x_4)+\kappa_2^2 (x_1+x_4)+\kappa_3^2(x_1+x_2)=\kappa^2(g_1+g_2)$$

This somewhat simplifies the system of equations.

In very special cases when eigenvalues are degenerate, you can double check if the Jordan form is the same (I don't know what happens to the conditions, if they remain independent or not).

• Thanks a lot!!! Commented Jan 16, 2019 at 8:05