Multiplying a matrix by its transpose

An earlier post claimed that there practical advantages to partitioning a matrix, thinking of the matrix as a matrix of matrices. This post will give an example.

LetM be a square matrix and suppose we need to multiply M by its transpose M^T. We can compute this product faster than multiplying two arbitrary matrices of the same size by exploiting the fact that MM^T will be a symmetric matrix.

We start by partitioningMinto four blocks

Then

and

Now for the first clever part: We don't have to compute both of the off-diagonal blocks because each is the transpose of the other. So we can reduce our calculation by 25% by not calculating one of the blocks, say the lower left block.

And now for the second clever part: apply the same procedure recursively. The diagonal blocks in MM^T involve a matrix times its transpose. That is, we can partitionA and use the same idea to compute AA^T and do the same for BB^T, CC^T, and DD^T. The off diagonal blocks require general matrix multiplications.

The net result is that we can compute MM^T in about 2/3 the time it would take to multiply two arbitrary matrices of the same size.

Recently a group of researchers found a way to take this idea even further, partitioning a matrix into a 4 by 4 matrix of 16 blocks and doing some clever tricks. The RXTX algorithm can compute MM^T in about 26/41 the time required to multiply arbitrary matrices, a savings of about 5%. A 5% improvement may be significant if it appears in the inner loop of a heavy computation. According to the authors, The algorithm was discovered by combining Machine Learning-based search methods with Combinatorial Optimization."

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• Embeddings, Projections, and Inverses
• The million dollar matrix multiply
• Vector spaces over finite fields

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