Re: Multiplication of matrices



On Sep 16, 10:58 am, mstem...@xxxxxxxxxxxxxxxxxxxx (Michael Stemper)
wrote:
I've spent several unproductive (if enjoyable) days trying to figure
out if a non-singular matrix can be written as the product of a singular
matrix and another matrix.

First, I came up with an algebraic proof that the product of a singular
and a non-singular matrix is always non-singular:

Let A, B be nxn matrices, with A singular and B non-singular, and let
AB = C. Then, ABB^-1 = CB^-1 = A. Since A is singular and B^-1 is
non-singular, and non-singular nxn matrices are closed under
multiplication, C must be singular.

I then tried to show that it was possible to have two singular matrices
whose product was non-singular. This was to be analgous to the demonstration
that (r-s) + s, with r rational and s irrational, was rational. Lacking
an inverse for singular matrices made this unfruitful.

Digging around, I found that the determinant of the product is the product
of the determinants (I think). This obviously would settle the issue, but
it's less than satisfying. I'd like an algebraic proof that it's not
possible to multiply two singular matrices and get a non-singular matrix.
Does such a proof exist?

Victor Meldrew has already suggested one approach. Here is another
one. The following are a couple of simply-proved facts about n x n
matrices. (1) If B is a nonsingular matrix, it transforms a set of n
linearly independent vectors v1, v2, ..., vn into linearly independent
vectors w1 = Bv1, w2 = Bv2, ..., wn = Bvn. Proof: B nonsingular
implies that no wi is the zero vector. If the wi are linearly
dependent, there exist scalars c1, c2, ..., cn, not all zero, such
that 0 = c1*w1 + ... + cn*wn = B*(c1*v1 + ... + cn*vn), so c1*v1 + ...
+ cn*vn = 0, contradicting the linear independence of the vi.
(2) If B is a singular matrix, it transforms a set of n vectors into
linearly dependent vectors. Obviously (and easily proved): if the vi
are dependent, so are the wi. However, the result asserts more: even
if the vi are linearly independent, the wi will be linearly dependent.
To prove this, assume the contrary. Then, the only solution to sum
ci*wi= 0 is c1 = c2 = ... = 0. Thus, the only solution to 0 = B*(sum
ci*vi) = 0 is c = 0. However, there is a nonzero solution to B*w = 0,
and the vi form a basis, so there are ci, not all zero, such that w =
sum ci*vi. This is a contradiction.

OK, so if C = A*B and B is singular, a set of n linearly independent
vectors is transformed by B into a linearly dependent set, and these
are in turn transformed into a linearly dependent set for any A,
singular or not. Thus, C transforms a basis into a linearly dependent
set, so is singular. Next, if B is nonsingular but A is singular, we
get a linearly independent set wi = B*vi, but then A then transforms
these into a linearly dependent set. Again, C is singular.

R.G. Vickson


To make it a little clearer, I'd like a proof that two elements of a
monoid with a sub-group cannot be multiplied to give an element of the
sub-group unless both of the elements are in the sub-group.

Any hints?

--
Michael F. Stemper
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Build a man a fire, and you warm him for a day. Set him on fire,
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.