# Square root of a complex symmetric matrix?

Is it possible to express a complex symmetric matrix $A$ as square of a matrix $B$ (i.e. $A = B^2$)? If $A$ were Hermitian, we could use Spectral Theorem to get $A = UDU^{-1}$ where $D$ has diagonal entries being eigenvalues of $A$. However, here $A$ being a complex symmetric seems to not give us any useful tools. The Takagi decomposition does not give us nice way to find a matrix for $A^{1/2}$ as well. Any ideas, suggestions?

Consider the matrix

$$\begin{pmatrix}i & 1 \\ 1 & -i \end{pmatrix}.$$

Its Jordan canonical form is

$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},$$

which has no square root. So it cannot have a square root since they are similar.

• If you have not yet encountered the Jordan form, you just need to know that the matrices are similar. The explicit similarity transform is given here under the heading 'Jordan decomposition'. – Philip Hoskins Apr 29 '16 at 6:40