numpy - Python: Calculating the inverse of a pseudo inverse matrix -

i trying calculate pseudo inverse of matrix should not difficult. problem inverting matrix.

i using following code:

a=numpy.random.random_sample((4,5,)) a=mat(a) b=pseudoinverse(a)  def pseudoinverse(a):     helper=a.t*a     print helper*helper.i     pi=helper.i*a.t     return pi` 

to test included print line. helper*helper.i should give unity. output is:

 [[ 2.    -1.     0.     0.     3.   ]   [ 0.     2.     0.     0.     3.5  ]   [ 0.    -0.5    1.125 -1.     2.25 ]   [ 2.     0.     0.25  -2.     3.   ]   [ 0.     0.     0.5   -2.     4.   ]] 

which not unity. don't know did wrong , know.

your matrix a not have full column rank. in consequence helper singular , not invertible (if print helper.i see large numbers).

the solution compute right inverse instead of left inverse:

helper = * a.t pi = a.t * helper.i 

see wikipedia more details.

unless doing exercise, use numpy's built in implementation of pseudeinverse.

edit

>>> numpy.random.seed(42) >>> = mat(numpy.random.random_sample((3, 4)))  # smaller matrix nicer output >>> h = * a.t >>> h * h.i matrix([[  1.00000000e+00,   1.33226763e-15,   0.00000000e+00],         [ -1.77635684e-15,   1.00000000e+00,   0.00000000e+00],         [  0.00000000e+00,   1.33226763e-15,   1.00000000e+00]]) 

up numeric precision looks pretty identity matrix me.

the problem in code a.t * a not invertible. if try invert such matrix wrong results. in contrast, a * a.t invertible.

you have 2 options:

1. change direction of multiplication
2. call pseudoinverse(a.t)