Thread: Invertible matrix is in same image as linear transformation

Let A be a 4x3 matrix and let B be a 4x4 matrix. Then, BA is a 4x3 matrix. Let L be the linear
transformation from R4 to R3 defined by L(x) = xA, and let M be the
linear transformation from R4 to R3 defined by M(x) = xBA. Now, assume that B is an invertible matrix so that C is a
4x4 matrix and CB=BC=Identity matrix.

Show that the image of L is the same as the image of M.

Attempt-

Let y be in the image of L so that y = xA.

Now, to get to y = xBA, I have to assume that x has an inverse, invert both sides, multiply each side by B, then C, and then move the x vector back over? I'm stuck.

Thanks

Hi, sfspitfire23.

I'll outline what came to my mind here; hopefully it will get things going in the right direction.

I think your idea of trying to express elements as images is the right one, we just need to apply it in a slightly different way. Since B is invertible, we know it is a matrix of full rank, i.e. rank(B)=4; this comes from the rank-nullity theorem (because an invertible matrix has only 0 in its kernel, i.e. its nullity is 0). Since B has full rank, for any , there is a unique such that . Then and the images are the same.

Does this help? Let me know if anything is unclear or in error.

Good luck!

Hmm, we havent studied rank yet, so I'm not sure that is needed.

suppose y is in Im(M).

then y = xBA = (xB)A = L(xB), so y is in Im(L). thus Im(M) is contained in Im(L).

on the other hand, suppose y is in Im(L), so y = xA. let z = xB^-1 (we can do this since B is invertible).

then M(z) = zBA = (xB^-1)BA = x(B^-1B)A = x(IA) = xA = y, so y is in Im(M). thus Im(L) = Im(M)

(the two sets contain each other).