Re: Proving matrices are singular.



I just showed you an example where A and B are *not* singular.
Therefore, it is false that they are singular in all cases under
those hypotheses.

-- m

I had to dig out my book to decide what I was trying to do in the
first place :P

What I'm trying to do is show that if Ax = Ay and x != y, that A must
be singular. Not A and B, just A. That changes things a bit. Let me
see if I can figure this out on my own....

I'm trying to show that if A is an nxn matrix and Ax = Ay and x != y,
A is singular (non-invertible)

Suppose that a is non-singular. This makes A invertible, and therefore
A^-1 * A = I.
So:

Ax = Ay
A^1 A x = A^1 A y
I x = I y
x = y

We see that if x = y, A must be non-Singular. Since x != y, then A is
singular.

Hmm...I'm not sure this is worded very well, would anyone care to
comment on this?
.

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