# Invertible matrix is in same image as linear transformation

• September 28th 2012, 06:49 AM
sfspitfire23
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
• September 28th 2012, 09:28 AM
GJA
Re: Invertible matrix is in same image as linear transformation
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 $x\in \mathbb{R}^{1\times 4}$, there is a unique $y\in \mathbb{R}^{1\times n}$ such that $x=yB$. Then $xA=yBA$ and the images are the same.

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

Good luck!
• September 28th 2012, 09:37 AM
sfspitfire23
Re: Invertible matrix is in same image as linear transformation
Hmm, we havent studied rank yet, so I'm not sure that is needed.
• September 28th 2012, 10:15 AM
Deveno
Re: Invertible matrix is in same image as linear transformation
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).