Question

In GNU Octave, How does matrix division work?

Instead of doing

1./[1;1]

I accidentally did

1/[1;1]

To my surprise this yields:

[0.5, 0.5]

The transverse case:

1/[1,1]

gives the expected:

error: operator /: nonconformant arguments (op1 is 1x1, op2 is 1x2)

Can someone explain the [0.5, 0.5] result?

Answer 1

this is a answer i got from Alan Boulton at the coursera machine learning course discussion forum:

The gist of the idea is that x / y is defined quite generally so that it can deal with matrices. Conceptually the / operator is trying to return x∗y−1 (or x * inv(y) in Octave-speak), as in the following example:

octave:1> eye(2)/[1 2;3 4]
ans =
  -2.00000   1.00000
   1.50000  -0.50000

octave:2> inv([1 2;3 4])
ans =
  -2.00000   1.00000
   1.50000  -0.50000

The trickiness happens when y is a column vector, in which case the inv(y) is undefined, so pinv(y), the psuedoinverse of y, is used.

octave:1> pinv([1;2])
ans =
   0.20000   0.40000

octave:2> 1/[1;2]
ans =
   0.20000   0.40000

The vector y needs to be compatible with x so that x * pinv(y) is well-defined. So it’s ok if y is a row vector, so long as x is compatible. See the following Octave session for illustration:

octave:18> pinv([1 2])
ans =
   0.20000
   0.40000

octave:19> 1/[1 2]
error: operator /: nonconformant arguments (op1 is 1x1, op2 is 1x2)
octave:19> eye(2)/[1 2]
ans =
   0.20000
   0.40000

octave:20> eye(2)/[1;2]
error: operator /: nonconformant arguments (op1 is 2x2, op2 is 2x1)
octave:20> 1/[1;2]
ans =
   0.20000   0.40000