## linear algebra proof

Linear spaces and subspaces, linear transformations, bases, etc.

### linear algebra proof

$A,B \in Mn(R)$ so that $AB=0n$ and $A$ is an invertible matrix. Proof that $B=0n$

by definition $A$ is invertible so:
$\exists C \in Mn : AC=CA=In$ so $A \ne 0n$
Then $AB=0n$ if $B=0n$
Here I can only say if and not if and only if because the product can be 0 even though both matrices are not 0.
I would like to know if I had it all wrong.
davide940

Posts: 2
Joined: Fri Apr 18, 2014 12:47 pm

davide940 wrote:$A,B \in Mn(R)$ so that $AB=0n$ and $A$ is an invertible matrix. Proof that $B=0n$

Please define your terms. What is "Mn(R)"? Is the "n" supposed to be subscripted? Does the "R" stand for "the set of real numbers"? What is "0n"? Is this perhaps an n-by-n zero matrix?

Thank you!

stapel_eliz

Posts: 1708
Joined: Mon Dec 08, 2008 4:22 pm