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I need to find the standard inverse of a large n*n matrix N (n=1,000,000).

I know that N*R=B and I have the matrix B, of the shape (n,k) (k=1,000).
I have the Moore-Penrose Inverse of matrix R (n,k).

I multiplied the B and the inverted R, and got a (n,n) matrix M. The sum of each row in M is very close to what the sum of each row in the standard inverse of N should be, but the actual values in M are not what I expect to have in N.

Please help me find the standard inverse of N.

dbank04
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  • $R R^+$ is not equal to identity in this case, as $n>k$. Only $R^+ R$ is equal to identity – Damien Dec 04 '18 at 13:49
  • The matrix $M$ has a rank equal to $k$, not $n$ – Damien Dec 04 '18 at 13:52
  • Damien, your comments are correct. Do you know a way in which I can get the standard inverse of N? – dbank04 Dec 05 '18 at 08:00
  • Your matrix is huge, and many standard algorithms exists to inverse a matrix. Difficult to advice a given algorithm without more knowledge on $N$: is it symmetric ? Is it sparse? Is it well or badly conditioned? – Damien Dec 05 '18 at 08:30
  • N is sparse and not symmetric. There are about 0.01% non-zero values in N. – dbank04 Dec 05 '18 at 08:43
  • You can make a research on 'inverting a sparse matrix'. For example here: https://math.stackexchange.com/questions/320302/computing-the-inverse-of-a-special-sparse-matrix – Damien Dec 05 '18 at 09:07

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