I am interested in the complexity of multiplying two matrices $A$ and $B$, i.e. to compute $AB$.

From [Le Gall and Urrotia], I know that:

  • if $A$ and $B$ are square-matrices of size $n$, then this can be done in $O(n^{\omega})$ where $\omega\approx 2.372$.
  • if $A$ has size $n\times n^{k}$ and $B$ has size $n^k \times n$ this can be done in $O(n^{\omega(k)})$ with $\omega(k)<\omega$ for $k<1$ (typically, $\omega(k)=2$ for $k\le0.3$).

I am wondering if there an efficient algorithm when $A$ has size $n^k\times n$ and $B$ has size $n\times n$ (or when $B$ has size $n\times n^{k}$), for $k\in(0,1)$.


  • when I say "efficient" algorithm, I mean an algorithm that a complexity smaller than the naïve $O(n^{2+k})$ algorithm. I am not refering to an actual implementation, just the existence of an algorithm.
  • I suppose that additing or multiplying two entries of a matrix is $O(1)$.

[Le Gall and Urrotia] Improved Rectangular Matrix Multiplication using Powers of the Coppersmith-Winograd Tensor by François Le Gall† Florent Urrutia

  • $\begingroup$ Are you assuming $k\in (0,1)$? $\endgroup$
    – kodlu
    Aug 25 '21 at 21:49
  • $\begingroup$ Indeed, I am only interested in the answer for $k\in(0,1)$. But the question remains valid for $k>1$ (and the cited paper provide bounds of $\omega(k)$ for $k>1$ as well. $\endgroup$
    – N. Gast
    Aug 26 '21 at 6:20
  • $\begingroup$ What counts as efficient? Anything less than the straightforward $n^k \times n \times n = n^{2+k}$? $\endgroup$ Aug 26 '21 at 7:21
  • $\begingroup$ Indeed, I was referring to anything with a complexity lower than the naïve complexity. I updated my question to reflect that. $\endgroup$
    – N. Gast
    Aug 26 '21 at 9:13

Assuming that efficient means better than the naive $O(n^{2+k})$ multiplication, let us review some possibilities.

Padding. For $k > \omega-2$, just pad $A$ with $n-n^k$ zero or garbage rows, perform square matrix multiplication in $O(n^\omega)$ time, and discard the extra rows from the output. This is already efficient because $\omega < 2+k$. (For $k < \omega-2$ the naive method would be better.)

Naive splitting. Following Knight (1995), we note that $\langle m,n,p \rangle$ matrix multiplication, with $m$ the smallest of the three dimensions, can be done in $O(m^{\omega-2}np)$ time. This is by splitting $A_{m \times n}$ into $1 \times (n/m)$ blocks of size $m \times m$, and splitting $B_{n \times p}$ into $(n/m) \times (p/m)$ blocks of same size. Then we naively perform the block matrix multiplication of size $\langle 1, \, n/m, \, p/m \rangle$ by performing $np/m^2$ square multiplications of size $m$, each in time $O(m^\omega)$, for a total of $O(m^{\omega-2}np)$. Applying to the current question with input size $\langle n^k, n, n \rangle$ we achieve $O(n^{k(\omega-2)}n^2) = O(n^{2+k(\omega-2)})$. This is efficient and better than padding for all $0<k<1$.

These are just some straightforward transformations that show that "efficient" is possible for the task given. I believe newer developments (Le Gall and others) may be provide more efficient methods.

Tensor powering. One thing to note is that the splitting method above used fast MM inside the square blocks, but naive MM to combine them. In fact one can use fast methods on both levels, and the blocks need not be square. Coppersmith and Winograd (1982) note that if we have two rectangular matrix multiplication methods,

  • $\langle m,n,p \rangle$ in $K$ multiplications and
  • $\langle m',n',p' \rangle$ in $K'$ multiplications,

then we can do

  • $\langle mm', nn', pp' \rangle$ in $KK'$ multiplications.

This is a powerful tool for combining MM methods of different dimensions. I am not sure, but possibly some of Le Gall & Urrutia's "symmetric" $\langle n, n^k, n \rangle$ rectangular methods that the OP cites could be "tensored up" in this way to solve the "asymmetric" $\langle n^k, n, n\rangle$ case for the current question.

Almost quadratic. Coppersmith (1982) reports that if $k = 0.17227$, then we can perform the $\langle n^k, n, n\rangle$ multiplication in $O(n^2 \, (\log n)^2)$ time, that is, almost in quadratic time! Note that the naive splitting would give an exponent of $2 + k(\omega-2) \le 2.0643$, using $\omega \le 2.373$.


Knight, Philip A., Fast rectangular matrix multiplication and QR decomposition, Linear Algebra Appl. 221, 69-81 (1995). ZBL0827.65044.

Coppersmith, D., Rapid multiplication of rectangular matrices, SIAM J. Comput. 11, 467-471 (1982). ZBL0486.68031.

Coppersmith, D.; Winograd, S., On the asymptotic complexity of matrix multiplication, SIAM J. Comput. 11, 472-492 (1982). ZBL0486.68030.

  • $\begingroup$ Thanks a lot for the references. It helps quite a lot. There is still a gap between 2+k(\omega-2) and \omega(k) but quite smaller than between \omega and \omega(k). $\endgroup$
    – N. Gast
    Aug 26 '21 at 9:18
  • $\begingroup$ In fact, according to sciencedirect.com/science/article/pii/S0885064X98904769, which cites a results of mathnet.ru/links/70088989aead859688ca36af8d30e329/rm5125.pdf (unfortunately in Russian), the complexity of $<n,n,n^k>$ is the same as the complexity of $<n,n^k,n>$, which is known as $\omega(k)$. $\endgroup$
    – N. Gast
    Aug 26 '21 at 12:45
  • $\begingroup$ Oh, cool! Indeed they say you can permute the three dimensions and keep the same complexity. I had a vague recollection of there being such a result but could not find it. $\endgroup$ Aug 26 '21 at 12:54

Thanks to Jukka's answer, I found the paper https://www.sciencedirect.com/science/article/pii/S0885064X98904769, which cites a result of http://www.mathnet.ru/links/70088989aead859688ca36af8d30e329/rm5125.pdf (unfortunately in Russian). If I understand this paper correctly (and in particular around Equation (2.7)), this paper claims that for any given exponent $r,s,t>0$, the complexity of the multiplication $<n^r,n^s,n^t>$ (i.e. multiply a $n^r\times n^s$ matrix by a $n^s\times n^t$) is independent by permutation of $r,s,t$.

This implies for my original question that the complexity of the multiplication $<n,n^k,n>$ is upper bounded by $\omega(k)$.

  • $\begingroup$ For bilinear matrix multiplication algorithms, you can find a slightly more informative statement of the index permutation property in V. Vassilevska Williams (2012), dl.acm.org/doi/10.1145/2213977.2214056 at the beginning of page 890. It relies on permuting the indices of the tensor that represents the algorithm (and observing that its tensor rank does not change). $\endgroup$ Aug 26 '21 at 13:06

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