# Is there more than one way to prove this?

1. Sep 23, 2012

### pyroknife

Suppose that A is an n × n matrix and u and v
are vectors in R^n. Show that if Au = Av and
u ≠ v, then A is not invertible.

This is the book's proof:
From the fact that Au = Av, we have A(u − v) = 0. If
A is invertible, then u − v = 0, that is, u = v, which
contradicts the statement that u = v.

Mine proof is, if A is invertible thus A^-1 exists
(A^-1)(Au)=(A^-1)(Av)
A^-1*A=I
→u=v
if u≠v then then the above does not hold, implying that A^-1 does not exist.

2. Sep 23, 2012

### gabbagabbahey

Your proof looks fine to me