Basis of Image of Linear Transformation

[lkh1986]

Homework Statement

Given a linear transformation F: R^3 --> R, F(x,y,z) = 3x-2y+z, find I am (F) and dim (Im (F))

Homework Equations

I have found that dim(ker F) = 2 and from the theorem dim (V) = dim (Ker F) + dim (Im F), I know dim (V) = 3, so dim (Im F) = 1.

The Attempt at a Solution

Im (F) = {F(x,y,z) | (x,y,z) inside R^3}
= (3x2y+z | x, y, z inside R}
= {R}
= L(1)

The vector (1) forms the basis for the image of F. Dim (Im F) = 1.

Is this the correct solution? It seems correct to me, since (1) is the standard basis for the vector space R.

Thanks.

 

[Dick]

Quite correct. You didn't even need to use the theorem. F(0,0,z)=z. It's pretty clear from that that Im(F)=R, right? And dim(R)=1.