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.

- Mar 8th 2013, 02:56 AMzachoonLinear 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. - Mar 8th 2013, 04:16 AMHallsofIvyRe: 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)? - Mar 8th 2013, 04:50 AMzachoonRe: 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. - Mar 8th 2013, 05:04 AMzachoonRe: 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.