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?



Strassen's algorithm
This $7$ is an absolute lower bound. The result is due to Hopcroft and Kerr "On minimizing the number of multiplications necessary for matrix multiplication." SIAM J. Appl. Math. (1971) and Winograd "On multiplication of $2\times 2$ matrices." Linear Algebra and Appl. (1971). [The former assume that entries of the matrices might not commute; while the latter gets the bound even assuming commutativity of the entries.] 


A lot more recently Landsberg showed that not only the rank but even the border rank of multiplication of $2 \times 2$ matrices is $7$, meaning very roughly that also small perturbations cannot lead to a smaller rank and thus saving of a multiplciation (for "approximate" calculations), assuming bilinearity of the algorithm. 


The paper establishing this is Landsberg "The border rank of the multiplication of $2\times 2$ matrices is seven", Journal Amer. Math Soc. 2006. The introduction also discusses your question.


See the link at the end for the respective volume of the journal, I think the article is free: <http://www.ams.org/journals/jams/2006-19-02/home.html> 


