# Mapping from n x n complex symmetric tridiagonal to 2n x 2n real symmetric tridiagonal

In my program I have a complex symmetric tridiagonal matrix. In order to perform some algorithmic optimizations I am searching for a (ideally linear) mapping from $n\times n$ complex symmetric tridiagonal to $2n \times 2n$ real symmetric tridiagonal. Does such a mapping exist?

## 1 Answer

I will assume you mean complex symmetric s.t. $A = A^T$ and not Hermitian $A = A^H$, which is different. Let $B \in \mathbb{R},[2n \times 2n]$ be your new tri-diagonal matrix. A simple option is each imaginary component is mapped $n$ places down the appropriate diagonal:

Main diag.: $b_{i,i} = Re(a_{i,i}), \;b_{i+n,i+n} = Im(a_{i,i}) \qquad \text{for} \qquad i = 1,\ldots,n$

Sub-diag.: $b_{i+1,i} = Re(a_{i+1,i}), \; b_{i+1+n,i+n} = Im(a_{i+1,i}) \qquad \text{for} \qquad i = 1,\ldots,n-1$

Super-diag.: $b_{i,i+1} = Re(a_{i,i+1}), \; b_{i+n,i+1+n} = Im(a_{i,i+1}) \qquad \text{for} \qquad i = 1,\ldots,n-1$