TITLE: Strassen's algorithm
QUESTION [7 upvotes]: I am reading Landsberg's "Tensors: Geometry and Applications". Here he mentions tensor formulation of Strassen's algorithm and shows that the rank of Strassen's matrix multiplication tensor is $7$ and $7$ is the lower bound for any such tensor and hence one needs $7$ multiplications for $2 \times 2$ matrix multiplication. My question is the following: is $7$ the absolute lower bound among all algorithms or just those that can be formulated via tensors?

REPLY [5 votes]: Quid's reply already gives you a complete answer for the $2\times 2$ case; however, there is an additional consideration that adds to the picture of the optimal exponent of matrix multiplication. 
One can prove that, for any $t$, if there exists a straight-line program that multiplies $n\times n$ matrices in $O(n^t)$ ops (without any bilinearity assumption or restriction on the operations made: divisions, additions, products of three more terms, everything is allowed), then there is a bilinear algorithm (i.e., one that can be "formulated with tensors") that achieves the same asymptotic complexity $O(n^t)$. So one does not lose generality by restricting to bilinear algorithms, when looking for the optimal exponent of matrix multiplication.
I had this same doubt myself a few years ago, and I found a proof of this statement in Borodin-Munro, The computational complexity of algebraic and numeric problems.