How Do Ranks of Composite Linear Transformations Compare?

[mathmathmad]

Homework Statement

Let S(U)=V and T(V)=W be linear maps where U,V, W are vector spaces over the same field K. Prove :

Homework Equations

a) Rank (TS) <= Rank (T)
b) Rank (TS) <= Rank (S)
c) if U=V and S is nonsingular then Rank (TS) = Rank (T)
d) if V=W and T is nonsingular then Rank (TS) = Rank (S)

The Attempt at a Solution

a) TS maps to W, so is T
b) TS maps to W, but S to V, but how do I show the ranks for (a) and (b)?
c) d) So inverse of S and T exists, and err...

U,V,W are vector space over the SAME field, does that mean they have the same number of entries, say R2, R3, etc etc

 

[HallsofIvy]

Homework Statement

Let S(U)=V and T(V)=W be linear maps where U,V, W are vector spaces over the same field K. Prove :

Homework Equations

a) Rank (TS) <= Rank (T)[/math]
The rank of a linear map is the dimension of its image space. Let w be in TS(U). Then there exist u in U such that TS(u)= w. Let v= S(u). Then T(v)= TS(u)= w. That is, if w is in TS(U) then it is in T(V). TS(U) is a subset of T(U) and, since they are both subspaces of W, TS(U) is a subspace of T(U).

b) Rank (TS) <= Rank (S)

Let $\{u_1, u_2,\cdot\cdot\cdot, u_n\}$ be a basis for U. Then $\{S(u_1), S(u_2), \cdot\cdot\cdot, S(u_n)\}$ spans S(U) (but is not neccessarily independent) so the dimension of S(U) is less than or equal to the dimension of U. Further $\{TS(u_1), TS(u_2), \cdot\cdot\cdot, TS(u_n)\}$ spans TS(U) but is not necessarily independent so dimension of TS(U) is less than or equal to the dimension of S(U).

c) if U=V and S is nonsingular then Rank (TS) = Rank (T)
d) if V=W and T is nonsingular then Rank (TS) = Rank (S)

The Attempt at a Solution

a) TS maps to W, so is T

I have no idea what you mean by this. What "is T"? Are you saying TS= T? That's obviously not true.

b) TS maps to W, but S to V, but how do I show the ranks for (a) and (b)?
c) d) So inverse of S and T exists, and err...

U,V,W are vector space over the SAME field, does that mean they have the same number of entries, say R2, R3, etc etc

I have no idea what you mean by "entries". If you mean dimension, no. R², R³, etc. are all "over the SAME field", the real numbers, but do NOT have the same "number of entries".

 

[mathmathmad]

"TS maps to W, so is T"

sorry, what I meant was T also maps to W

for (c) and (d)
what does the the nonsingularity of the matrix imply?
how does that show that the ranks are both equal?

 

[vintwc]

"TS maps to W, so is T"

sorry, what I meant was T also maps to W

for (c) and (d)
what does the the nonsingularity of the matrix imply?
how does that show that the ranks are both equal?

I think the best way to show that the ranks are equal is by considering the surjectivity and injectivity of S and T respectively (since S and T are non-singular). From there, it should be pretty much straightforward.