# Matrix multiplication

Let I(n) and U(n) be the number of steps needed to invert an nxn matrix and nxn upper triangle matrix respectively. Can we prove I(n)<=cU(n), where c is some constant?

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The answer is probably No, if you wish a constant independent of $n$. On the one hand, the naive method gives the optimal result $U(n)=n^2$. On the other hand, it is known that the complexity of inversion and that of matrix multiplication are the same (see for instance the second edition, to appear soon, of my book Matrices;Theory and Applications, GTM 216 Springer-Verlag, 2010). If the answer to your question is positive, this implies therefore that matrix multiplication can be done in $O(n^2)$ operations. This is highly unlikely. The state of the art tells us that it can be done in $O(n^{2.376})$ operations. Optimists believe that it could be done in $O(n^{2+\epsilon})$ for every $\epsilon>0$, but not in $O(n^2)$.
@Matt. Conversely, inversion of $3n\times 3n$ matrices can be used to multiply $n\times n$ matrices with the same complexity, up to a universal constant. If $A$ and $B$ are given, just invert the block triangular matrix whose diagonal is ($I_n$ $I_n$ $I_n$) and is boardered by the diagonal ($A$ $B$); the other blocks are $0_n$'s. –  Denis Serre Sep 17 '10 at 13:37
Denis, I'm confused. To me, the naive method of inverting an upper triangular matrix has complexity $n^3$. Moreover, your answer seem to show that the answer is YES: You have just shown that $U(3n)$ bounds $n \times n$ matrix multiplication. –  David Speyer Jul 11 '12 at 14:14
@David. You're right. I confused between solving $Ux=b$ and inverting $U$. –  Denis Serre Jul 11 '12 at 14:40