Suppose U and V are finite-dimensional vector spaces and S∈L(V,W) and T∈L(U,V). Prove that
dimnull ST≤dimnull S+dimnull T
It sufficies to show
dimnull ST=dimnull S∣range T+dimnull T
Since null S∣range T⊆null S makes the inequality obvious
dimnull ST≤dimnull S+dimnull T
In order to prove dimnull ST=dimnull S∣range T+dimnull T I'm going to construct u1,…,um which is a basis for null ST such that u1,…,ur is a basis for null T and Tur+1,…,Tum is a basis for null S∣range T.
Let u1,…,ur be a basis for null T then extend it to a basis u1,…,um of null ST. It suffices to show Tur+1,…,Tum is a basis for null S∣range T. It's independent because if
ar+1Tur+1+⋯+amTum=0
Then (by linearity) ar+1ur+1+⋯+amum∈null T meaning we can write
b1u1+⋯+brur=ar+1ur+1+⋯+amum
Independence of u1,…,um implies each bj=aj=0. Thus Tur+1,…,Tum is independent, now it suffices to show it spans null S∣range T.
Let v∈null S∣range T. since v∈range T there is some u∈U such that v=Tu, and since STu=0 we can write
u=c1u1+⋯+cmum
Apply T to both sides and use the fact that Tuj=0 for 1≤j≤r
v=cr+1Tur+1+⋯+cmTum
Which shows that Tur+1,…,Tum spans null S∣range T completing the proof that it is a basis.
I don't like how long it took to prove null ST=null S∣range T+null T. todo: look for a shorter proof by starting with a basis for null S∣range T
This was intuitively obvious but annoying to prove rigorously. My first solution (shown below) is ugly and I think wrong. I'll leave it here though since it's instructive
Old (bad) Solution
Let u1,…,ur be a basis for null T, since null T⊆null ST we can extend it to a basis u1,…,um of null ST. Let R=span(ur+1,…,um). T∣R (T restricted to R) is injective since
null T∣R=null(T)∩R=span(u1,…,ur)∩span(ur+1,…,um)={0}
Therefor Tur+1,…,Tum are linearly independent in V meaning we can extend it to a basis of V
Tur+1,…,Tum,v1,…,vn
which implies dimnull ST≤dimnull S+dimnull T since we constructed a basis for null S and null T by extending a basis of null ST.