[Strassen Algoritm](http://en.wikipedia.org/wiki/Strassen_algorithm) is a well-known matrix multiplication divide and conquer algorithm.
The trick of the algorithm is reducing the number of multiplications to 7 instead of 8. I was wondering, can we reduce any further? Can we only do 6 multiplications?


Also, what happens if we divide the NxN arrays into 9 arrays each of (N/3)x(N/3) instead of 4 arrays of (N/2)x(N/2). Can we then do less multiplications?



Strassen Algorithm 7 multiplications
You may be interested to know that there's a way to multiply $3\times3$ matrices using only 23 multiplications (where the naive method uses $27$). See Julian D. Laderman, A noncommutative algorithm for multiplying $3\times3$ matrices using $23$ muliplications, Bull. Amer. Math. Soc. 82 (1976) 126–128, MR0395320 (52 #16117). 


As for doing $2\times2$ with fewer than $7$ multiplications, this was proved impossible just a few years ago. See J M Landsberg, The border rank of the multiplication of $2\times2$ matrices is seven, J. Amer. Math. Soc. 19 (2006), 447–459, MR2188132 (2006j:68034). 


EDIT: As Mariano points out, Landsberg acknowledged a gap in the proof. But don't panic. The review, and my preceding paragraph, were based on the electronic version of Landsberg's paper. The print version (which is freely available on the AMS website) is different. It says, "Hopcroft and Kerr [12] and Winograd [22] proved independently that there is no algorithm for multiplying $2\times2$ matrices using only six multiplications." 


Those references are 


J. E. Hopcroft and L. R. Kerr, On minimizing the number of multiplications necessary for
matrix multiplication, SIAM J. Appl. Math. 20 (1971), 30–36, MR0274293 (43:58). 


S.Winograd, On multiplication of $2\times2$ matrices, Linear Algebra and Appl. 4 (1971), 381–388, MR0297115 (45:6173). 


