- #1

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I'm not seeing (perhaps because I don't want to ;)) how the above is not always true. Can anyone show me an example proving me wrong?

Thanks!

- Thread starter sust0005
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- #1

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I'm not seeing (perhaps because I don't want to ;)) how the above is not always true. Can anyone show me an example proving me wrong?

Thanks!

- #2

Mark44

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If you are given that BC = 0, multiply on the left of both sides by B

- #3

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- #4

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haha, I've seen thru the cloud of trying to convince myself I was correct. My error is now obvious!

- #5

Mark44

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For example, consider these matrices B and C.

[tex]B~=~\left[\begin{array}{c c}

0&1\\

0&0

\end{array}\right]

[/tex]

[tex]C~=~\left[\begin{array}{c c}

0&2\\

0&0

\end{array}\right]

[/tex]

For these matrices BC = 0, yet neither B nor C is the zero matrix.

- #6

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This is fairly common, so you should try to keep it in mind. It pops up in a lot of places; matrices are just one example. For instance, the ring of integers modulo 6 is not an integral domain since 2 and 3 are both nonzero but their product is 0 modulo 6. One can also find positive powers of nonzero elements that are zero. Once again using modular arithmetic for an example, the square of two is congruent to 0 modulo 4.

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