Dim and null question

1. Feb 13, 2008

rjw5002

Ok. i have a problem that states: "suppose U and V are finite-dimensional vector spaces and that S $$\in$$ L(V,W), T $$\in$$ L(U,V). prove that
dim(null(ST))$$\leq$$ dim(null(S)) + dim(null(T)).

i think that some call the null space the kernel.

i have tried using the fact that null(S) is a subspace of V, null(T) is a subspace of U, and null(ST) is a subspace of U.
also, that:
dim(null(T)) $$\geq$$ dim(U) - dim(V)
dim(null(S)) $$\geq$$ dim(V) - dim(W)
dim(null(ST)) $$\geq$$ dim(U) - dim(W)

the closest that i get is that dim(null(ST))= dim(null(S)) + dim(null(T)).

Can anyone offer any suggestions?

2. Feb 13, 2008

John Creighto

It's been a while since I took a linear algebra course but maybe it would be helpful to assign a variable call it N to the total number of dimensions.

BTW what does dim(nul(ST)) mean?

3. Feb 13, 2008

Hurkyl

Staff Emeritus
I hope you didn't get that, because it's patently false.
For example, let U,V,W = R^2, and let S,T both be the matrix
[1, 0]
[0, 0]

You've thought about dimension and you've thought about nullity... is there any related quantity you haven't thought about?

4. Feb 13, 2008

John Creighto

So I still want to know what ST is. Is it the inner product, the Cartesian product or is it a matrix product?

5. Feb 13, 2008

Hurkyl

Staff Emeritus
Composition of functions.

6. Feb 13, 2008

John Creighto

Why should that be the case? I don't see the words composition of functions anywhere in the definition of a vector space.

7. Feb 14, 2008

HallsofIvy

Staff Emeritus
You also won't see "linear transformation" anywhere in the definition of vector space!

After you have vector spaces, you can define linear functions on them and the composition is a natural operation on those functions.

The original post specifically that T was a linear transformation from U to V and that S was a linear transformation from V to W. The composition, ST, is a linear transformation from U to W.

If S and T are represented as matrices, in some appropriate bases for U, V, and W, then ST would correspond to the matric multiplication.

8. Feb 14, 2008

rjw5002

yes, sorry if that was unclear. ST is meant to be the composition of the functions.

9. Feb 14, 2008

HallsofIvy

Staff Emeritus
Suppose v is in the null space of T. Then Tv= 0 so STv= 0 and v is in the null space of ST. That shows that the null space of T is a subspace of the null space of ST. Now suppose v is in U but NOT in the null space of T. Then Tv is a non-zero vector in V and, in order that STv=0, must be in the null space of T. That is, v is in T-1(null space of V). That set cannot have dimension larger than dim(null space of S). Putting those together, the dimension of the null space of TS cannot be larger than the dim(null space of S)+ dim(null space of T).

10. Feb 14, 2008

mathwonk

the null space of S is the space of stuff going to zero, so the null space of ST is the stuff S sends into the null space of T.

so we need to estimate the size of the pullback K of null(T) under S. this follows from the basic formula relating dimensions of nullspace and image.

i.e. the image of K is inside null(T), so that formula says the dimension of K is at most the sum of dim null(T) and dim null(S).

(recall the fundamental formula says that dim im(U) + dim null(U) = dim(dom(U)).

apply this to S acting on K as domain, and then the image is contained in null(T).))

11. Feb 14, 2008

HallsofIvy

Staff Emeritus
Mathwonk, I was with you until you swapped T for U:
"(recall the fundamental formula says that dim im(U) + dim null(U) = dim(dom(U))."

In the original question, U was a vector space, not a linear transformation!