If BC = 0 where B is invertible, show C = 0

[sust0005]

Isn’t it true that if B is invertible then B DNE 0, and if B DNE 0 then the only way to have BC = 0 is to have C = 0?

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!

 

[Mark44]

If B is invertible, then B^-1 exists.

If you are given that BC = 0, multiply on the left of both sides by B^-1. What do you get?

 

[sust0005]

Yes, that I understand (and wish I would've put down on the final!), but what I don't understand is why my initial statement is untrue. That is, why is that not a valid proof that C = 0?

 

[sust0005]

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

 

[Mark44]

Because you didn't use the given information that B is invertible.
For example, consider these matrices B and C.
$$B~=~\left[\begin{array}{c c} 0&1\\ 0&0 \end{array}\right]$$

$$C~=~\left[\begin{array}{c c} 0&2\\ 0&0 \end{array}\right]$$

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

 

[zpconn]

Rings of matrices are not in general integral domains (even if the elements are drawn from a field), as Mark44 illustrated with an example. Thus, if you know that BC = 0 and B != 0, you can't conclude that C must be 0 as you would in an integral domain.

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.