# When is (S o T) invertible? (Linear Maps)

1. May 21, 2012

### JoshMaths

1. The problem statement, all variables and given/known data
Show SoT is invertible iff
kerT = {0}
ImS = W
ImT n kerS = {0} and ImtT (+) kerS = V

T: U -> V
S: V -> W

3. The attempt at a solution

We know if kerSoT = {0} and kerT C kerSoT then kerT = {0}

ImS = W is implying surjectivity?
SoT = S(Tu) = Sv = W = ImS for all v (belongs to) ImT

I know the last part is stating the rank-nullity theorem but in a way I have not seen. If the kernel of S is zero then doesn't ImT = V or is kerS not necessarily zero?

I am also have difficulty with the whole premise of using set notations to explain properties of linear maps so if you could provide some guidance on this that would be great.

Josh

2. May 21, 2012

### vela

Staff Emeritus
Depends on which mapping you're talking about. It tells you S is surjective, but S○T may not necessarily be. For example, if T mapped everything to 0, the image of S○T would be {0}.

Your notation is a mess. You're equating a mapping (S○T) to vectors (S(Tu) and Sv) to a vector space (W).

Suppose S: ℝ3→ℝ2 maps (x,y,z) to (x,y), and T: ℝ2→ℝ3 maps (x,y) to (x,y,0). The kernel of S is the z-axis, but S○T is invertible.