在Octave / Matlab中pinv（[inf]）= NaN的方法

Question

I am using Octave 3.8.1, a Matlab-like program. I'd like to generalize 1/x to the case where x may be a scalar or a matrix. Replacing 1/x with inv(x) or pinv(x) works for most x , except:

octave:1> 1/inf
ans = 0

octave:2> pinv([inf])
ans = NaN

octave:3> inv([inf])
warning: inverse: matrix singular to machine precision, rcond = 0
ans = Inf

Should I convert NaN to 0 afterwards to get this to work? Or have I missed something? Thanks!

Answer 1

The Moore–Penrose pseudo inverse , which is the basis for Matab and octave's pinv , is implemented via completely different algorithm than the inv function. More specifically, singular value decomposition is used , which require's finite-valued matrices (they also can't be sparse ). You didn't say if your matrices are square or not. The real use of pinv is for solving non-square systems ( over- or underdetermined ).

However, you shouldn't be using pinv or inv for your application, no matter the dimension of your matrices. Instead you should use mldivide ( octave , Matlab ), ie, the backslash operator, \\ . This is much more efficient and numerically robust.

A1 = 3;
A2 = [1 2 1;2 4 6;1 1 3];
A1inv = A1\1
A2inv = A2\eye(size(A2))

The mldivide function handles rectangular matrices too, but you will get different answers for underdetermined systems compared to pinv because the two use different methods to choose the solution.

A3 = [1 2 1;2 4 6]; % Underdetermined
A4 = [1 2;2 4;1 1]; % Overdetermined
A3inv = A3\eye(min(size(A3))) % Compare to pinv(A3), different answer
A4inv = A4\eye(max(size(A4))) % Compare to pinv(A4), same answer

If you run the code above, you'll see that you get a slightly different result for A3inv as compared to what is returned by pinv(A3) . However, both are valid solutions.