Strassen algorithm
Strassen algorithm for a symmetric matrix:
For a square matrix [math]\displaystyle{ A \in \mathbb{C}^{n \times n} }[/math] where [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] is about [math]\displaystyle{ 2/3 }[/math] 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{ A^TA = \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] and [math]\displaystyle{ T_q(2n) = 4T_q(n) + 2n^{(_2 7)} }[/math] or [math]\displaystyle{ T_q(n) = 4T_q(n/2) + 2/7n^{(_2 7)} }[/math].
The geometric series yields because of [math]\displaystyle{ T_q(1) = 1 }[/math]: [math]\displaystyle{ T_q(n) = 64T_q(n/8) + 2/7n^{(_2 7)}(1+4/7 + (4/7)^2 + ...) = 2/7n^{(_2 7)} (1-(4/7)^{(_2n)})/(1-4/7) = 2/3n^{(_2 7)}.\square }[/math]