# Thread: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

1. ## Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

Let U, V and W be finite-dimensional vector spaces over F and S is in L(V,W) and T is in L(V,W). Show that:
nullity(composite of f and g) is less than or equal to nullity(T)+nullity(S).

Thank you.

2. ## Re: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

Frankly, you seem to have very little interest in these problems- you are not even copying them correctly! (Or your text book is just really really bad.) You cannot talk about "nullity of f and g" when there are NO functions called "f" or "g" in the problem. Since you do talk about linear functions S and T, I would think that you meant "the composition of S and T" but that still makes no sense. S is in L(V, W) so S maps vectors in V to vectors in W. T is also in L(V, W) so, again, maps vectors in V to W. To be able to compose ST, we would have to have T in L(V, W) and S in L(W, Z) for some vector space Z. To be able to compose TS, we would have to have S in L(V, W) and T in L(W, Z).

If you meant "S is in L(V, W) and T is in L(W, Z), show that nullity(TS) is less than or equal to nullity(T)+ nullity(S)", then take v in nullity(S). What is S(v)? What is TS(v)?

3. ## Re: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

Oh whoops my bad. I'm a bit out it. here, let me copy it correctly.

Let U, V and W be finite-dimensional vector spaces over F and S is in L(U,V) and T is in L(V,W). Show that:
nullity(composite of T and S) is less than or equal to nullity(T)+nullity(S).

So yes, you had the right idea in interpretting my mis-copied content.

I'm not quite sure, but i had the idea of starting the proof off with stating that since null(S) is a subspace of U and null(T) is a subspace of T, null(S) has a basis (u1, u2, ..., un), and null(T) has a basis (v1, v2, ..., vm), thus implying that nullity(S)=n and nullity(T)=m. However, i'm not quite sure what the relation is between compositions and nullities...

thank you for the response.

4. ## Re: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

*Correction, i'm not quite sure what the relation is between compositions, nullities, and dimensions, meaning i don't understand how to put these ideas together. I understand that the composition is T(S(u)) given that u is in U. But from that point on, i'm not sure what to do or how to think about the problem.

5. ## Re: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

Hi! I'm new here and I'm having almost the same problem as this, If V and W are finite dimentional over f. How can we show that If g:V->W and h:W->U are linear transformation in F, then the nullity of the composite (h and g) is less than or equal to the nullity of h + the nullity of g?

Thanks!

6. ## Re: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

A good start to any problem like this is a review of the definitions of the words you are using!

The "null space" of a linear transformation, T, from U to V is the set of all vectors, u, in U, such that T(u)= 0, the 0 vector in V. It is fairly easy to show that the "null space" is a subspace of U and so has some integer dimension. The "nullity" of a linear transformation is the dimension of null space.

Suppose T, a linear transformation from U to V, has null space X, of dimension n, and that S, a linear transformation from U to W, has null space Y, with dimension m. If x, in U, is in the null space of ST, then T(x) must be in the null space of S. Of course, T(0)= 0 so any x in the null space of T is a subspace of that set. Now what if T(x) is not 0 but T(x) is in the null space of S?

7. ## Re: Linear Algebra Proofs regarding Linear maps, null spaces, and composition.

Originally Posted by zachoon
Oh whoops my bad. I'm a bit out it. here, let me copy it correctly.

Let U, V and W be finite-dimensional vector spaces over F and S is in L(U,V) and T is in L(V,W). Show that:
nullity(composite of T and S) is less than or equal to nullity(T)+nullity(S).

So yes, you had the right idea in interpretting my mis-copied content.

I'm not quite sure, but i had the idea of starting the proof off with stating that since null(S) is a subspace of U and null(T) is a subspace of T, null(S) has a basis (u1, u2, ..., un), and null(T) has a basis (v1, v2, ..., vm), thus implying that nullity(S)=n and nullity(T)=m. However, i'm not quite sure what the relation is between compositions and nullities...

thank you for the response.
null(S) has a basis $\displaystyle \left\{u_1,u_2,\text{...},u_n\right\}$

expand this basis to a basis of null(TS) say $\displaystyle \left\{u_1,u_2,\text{...},u_n,u_{n+1},\text{...},u _r\right\}$

now consider the vectors $\displaystyle \left\{S\left(u_{n+1}\right),\text{...},S\left(u_r \right)\right\}$ in V