Strassen algorithm
Strassen algorithm for a symmetric matrix:
For a square matrix [math]\displaystyle{ A \in \mathbb{C}^{n \times n} }[/math] where sufficiently big [math]\displaystyle{ 2^k := n, k \in \mathbb{N}^* }[/math], the runtime [math]\displaystyle{ T_q(n) }[/math] of the Strassen algorithm for the matrix product [math]\displaystyle{ AA^T }[/math] roughly halves that of the original algorithm in [math]\displaystyle{ \mathcal{O}(n^{(_2 7)}) }[/math].
Proof: For [math]\displaystyle{ A := \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} }[/math], it holds that [math]\displaystyle{ AA^T = \begin{pmatrix} A_{11}A_{11}^T+A_{12}A_{12}^T & A_{11}A_{21}^T+A_{12}A_{22}^T \\ A_{21}A_{11}^T+A_{22}A_{12}^T & A_{21}A_{21}^T+A_{22}A_{22}^T \end{pmatrix} }[/math].
The geometric series yields the claim because of the halvings above (see top right).[math]\displaystyle{ \square }[/math]