Short question about the homogenous system

Question

I am working on a text book problem for a intro linear course. But the solution is not in the back. I am looking to see if I understand it correctly.

The question asks, " If A is a matrix, and the system AX=O has a non-trivial solution, show that there is no matrix, B, such that BA=I."

I am not sure if what Im thinking is correct. If AX=O has a non-trivial solution, this implies that there is then infinite solutions ( because of parameters). And now I am not sure where to go. I tried writing it like BA=I, (BA=I)I, (BA)I=I^2, but all this gets me is that I=I..

Thanks a lot for any suggestions/help.

Answer 1

If $V$ is a nontrivial solution and $BA=I$, then $$ V = IV = (BA)V = B(AV) = BO = O, $$ which contradicts $V$ being nontrivial.

Answer 2

HINT:

$A$ multiplied on the right by $X$ gets $0$ and multiplied on the left by $B$ gets $I$. Multiply on both sides and use associativity, that is, you can first multiply on the right by $X$, then on the left by $B$, you get the same as: first multiply on the left by $B$ then multiply on the right by $X$.

Answer 3

Not sure if this is what you were looking for, but if you are looking for the intuition behind this idea this little explanation might help.

We have a matrix A such that the equation Ax = 0 has a non-trivial solution. So what you say is correct, there will be a free-variable (at least one) and we will have infinite solutions to the equation.

To aid your understanding, think of the matrix A as something that operates on the vector x. I don't know if you've read about linear transformations yet, but one way to think of the equation Ax = b, is that a matrix A is acting on a vector x and giving you a vector b. So a linear transformation is basically a function that acts on a vector to return a vector.

Now, if the equation Ax = 0 has more than one solution, this means that the function is mapping more than one vector to the vector 0. Therefore, the function is not one-one. If you have studied functions before, you will have probably read that a one-to-one function always has an inverse function. Because each vector in your domain goes to precisley one vector in your codomain, you can always go the other way. Therefore, there would be an inverse matrix that you could multiply by b, to get back x. If you called this matrix B, then BA = I, the identity matrix.

In the case of this problem, the transformation is not one-to-one. So given the output vector 0, you can't tell me which vector x was input to get there. Therefore, you can't find an inverse matrix that will take you from b to x, or when multiplied by A will give you the identity matrix.

In fact, there is a method for finding inverse matrices that involves creating the augmented matrix [A|I] and row reducing until you get [I|A^-1]. If at the end you can't get I on the left side, there is no inverse matrix for A. There are other methods as well but that is one to try.