Every square complex matrix is similar to a TRIDIAGONAL complex-symmetric matrix?

Question

Every square complex matrix is similar to a complex-symmetric matrix. But I think a stronger statement is also true: Every such matrix is also similar to a TRIDIAGONAL complex-symmetric matrix. Where is this proved? How can it be proven?

Define a complex-symmetric matrix $M$ to be one where $M = M^T$. Define it to be tridiagonal if $m_{ij}=0$ whenever $|i - j| > 1$.

Empirically, the statement is easy to verify by finding t.c.s (tridiagonal complex-symmetric) matrices similar to the Jordan block $J_n(0)$ of size up to at most $n=7$ of eigenvalue $0$. There are multiple t.c.s matrices similar to any $J_n(0)$ though, and it's hard to pick a "simplest" one. I think one method of proof might involve stuff related to the Lanczos algorithm?

Answer 1

With reference to Wikipedia, take this algorithm: https://en.wikipedia.org/wiki/Lanczos_algorithm#A_more_provident_power_method

I refer to the last algorithm with a numbered list of instructions.

Instruction number 5 must be adapted. Of course, by $||v||$ we really mean $\sqrt{v^T v}$ in this case.

In this case:

• If $h_{j,j+1} \neq 0$ then let $v_{j+1} = w_{j+1}/h_{j,j+1}$.

• If on the other hand we have $w_{j+1} = 0$, choose an arbitrary unit vector $v_{j+1}$ orthogonal to $v_1, \dotsc, v_j$. This exists by Witt's embedding theorem.

• If on the other hand we have $w_{j+1} \neq 0$ and $||w_{j+1}||=0$ which is the remaining case, choose a non-zero $v_{j+1} \in (\operatorname{span}(v_1, v_2, \dotsc,v_j)^\perp \cap \operatorname{span}(v_1, v_2, \dotsc,w_{j+1})) \setminus\operatorname{span}(w_{j+1})$, where we've used set subtraction. It's possible to verify that such a choice exists and results in $||v_{j+1}|| \neq 0$. Normalise this $v_{j+1}$ and then set $h_{j,j+1}$ to be the coefficient of $v_{j+1}$ when $Av_j$ is expressed as a linear combination of $\{v_1, v_2, \dotsc, v_{j+1}\}$.

In this complex-orthogonal basis, $A$ is Hessenberg and complex-symmetric, and therefore also tridiagonal.