If $A$, $B$, $C$, $D$ are arbitrary matrices in $\mathrm{GL}(n, \mathbb{R})$, prove $AB\neq CD$.

Suppose I consider the set of all matrices in $\mathrm{GL}(n, \mathbb{R})$, and I arbitrarily pick four distinct matrices $A,B,C,D$.

How can one prove that $AB$ is not equal to $CD$.

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They could be equal unless you have further information. For example, if $n \geq 2,$ it is easy to find matrices $A,B,C,D,$ each of order $3,$ and all different, so that $AB = CD = I.$ – Geoff Robinson Mar 9 '12 at 18:25
Suppose AB or CD cannot be equal to I – user996522 Mar 9 '12 at 18:48
That's not restrictive enough. Let $A = I$, $B = 6I$, $C = 2I$, $D = 3I$. – Adam Saltz Mar 9 '12 at 19:28
@user996522 : That wa just one easy example. You need to have much more information befoe you can be sure that $AB \neq CD$. – Geoff Robinson Mar 9 '12 at 19:32
You can't prove it, because it's not true in general. It's possible that you might argue, with a lot of machinery to make sense of things, the you will "generally" have $AB\neq CD$ (that is, that under a suitable definition of "arbitrarily pick four distinct matrices" which allows a real probabilistic interpretation, the probability that they are equal is extremely small; though I'm not sure if that cna be made sensible), but you can't prove it always happens, because it doesn't. For instance, if $A$, $A^2$, $A^3$, and $A^4$ are pairwise distinct, you still get $AA^4 = A^2A^3$. – Arturo Magidin Mar 9 '12 at 19:38

Note that $AB=CD$ is equivalent to $D= C^{-1}AB$. If you pick four matrices at random, there are infinitely many choices for $D$ but only one of them will fail
$$AB \neq CD \,.$$
This shows that $AB \neq CD$ almost surely.