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I need to prove or disprove whether non-singular matrices commute. Can you help me and tell me how? What should i use cos i really don't know, I am beginner in this.

Thanks in advance.

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13  
Make up two random two-by-two matrices, multiply in both ways, and report back. –  J. M. Nov 6 '10 at 21:49
2  
The best thing to do when you're not sure what to do is try it out. –  Yuval Filmus Nov 6 '10 at 22:57
    
I agree with both of the two comments above. If the OP is looking for more courage: in fact two nonsingular matrices, chosen at random, "usually" do not commute, so probably the first example you try will work. Just make sure not to take either of them of the form $[a 0][0 a]$ -- such matrices commute with every matrix. –  Pete L. Clark Nov 6 '10 at 23:13

1 Answer 1

This is one of those problems where one example that contradicts the claim is enough to disprove it (also known as giving a counterexample).

The phrase non-singular matrices indicates we are talking about square matrices, and presumably you want them to be of equal size (since otherwise their product is not defined in either order).

So as the commentators have suggested, we should try a couple of "random" two-by-two matrices, as these will "usually" not commute.

Here are a pair of non-commuting non-singular matrices:

1  0
0 -1

1  1
0  1

Multiplying by the first of these matrices on the left scales the second row by a factor of -1 (and otherwise leaves the entries of the first row unchanged), while multiplying by it on the right scales the second column by a factor of -1 (and otherwise leaves the entries of the first column unchanged).

Try and you'll quickly see that the pair of matrices above do not commute (but both are non-singular, e.g. because their determinants are nonzero).

regards, hm

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5  
-1 for doing someone's homework for them when there is no indication that they could not do it for themselves. –  Pete L. Clark Nov 7 '10 at 0:39

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