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

  • Thread starter sust0005
  • Start date
  • #1
3
0
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!
 

Answers and Replies

  • #2
33,719
5,417
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?
 
  • #3
3
0
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?
 
  • #4
3
0
haha, I've seen thru the cloud of trying to convince myself I was correct. My error is now obvious!
 
  • #5
33,719
5,417
Because you didn't use the given information that B is invertible.
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
243
2
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.
 

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