Question

Let dim V = n. Prove that if T ∈ L(V ), then V = nullT n ⊕ rangeT n.​

Answer 1

Answer:

Let's prove something slightly more general.

Assume that V is a finite dimensional vector space and T∈L(V). We have the nested chain of ascending subspaces

{0}=ker(T0)⊆ker(T)⊆ker(T2)⊆…

and since V is finite dimensional, this chain must stabilize after finitely many steps. In fact, it must stabilize after at most n steps (where n=dimV). Let N∈N0 be the minimal number such that ker(TN)=ker(TN+1) (so N≤n). Then the chain of subspaces looks like

{0}=ker(T0)⊊ker(T1)⊊⋯⊆ker(TN)=ker(TN+1)=ker(TN+2)=…

We also have the nested chain of descending subspaces

V=im(T0)⊇im(T)⊇im(T2)⊇…

Since dimker(Ti)+dimim(Ti)=n, this chain also stabilizes precisely after N steps so we also have

V=im(T0)⊋im(T)⊋⋯⊋im(TN)=im(TN+1)=im(TN+2)=…

Let's show that V=ker(TN)⊕im(TN). Since dimker(TN)+dimim(TN)=n, it is enough to show that ker(TN)∩im(TN)={0}. Let TN(v)∈ker(TN)∩im(TN) so TN(TN(v))=T2N(v)=0. Since 2N≥N, we get v∈ker(T2N)=ker(TN) so TN(v)=0. Note that since the chains stabilize after N steps, we also have

V=ker(Ti)⊕im(Ti)

for all i≥N.

Now let's get back to your problem. If the chains above stabilize at N=n then

n≥dimker(Tn)>dimker(Tn−1)>⋯>dimker(T0)=0

which implies that in fact dimker(Tn)=n so Tn=0 and T is nilpotent. Hence, if T is not nilpotent, N≤n−1 and so we have

V=ker(Tn−1)⊕im(Tn−1).