Answer to Question #175119 in Linear Algebra for Sonu

Question

Let T1 and T2 be linear operators on V such that the range of T2 is contained in the kernel of T1.

Show that T1 ◦ T2 = 0. Also show that rank(T1)+rank(T2) ≤ dim(V )

Answer 1

Let T₁ and T₂ be linear operators on V such that the range of T₂ is contained in the kernel of T₁.

$\implies$ T₂(x) $\subseteq$ ker(T₁)

since we know that if a$\in$ ker(T₁) then T₁(a) = 0_v , where 0_v is additive identity of V

$\therefore$ T₁(T₂(x)) = 0_v $\forall$ x$\in$ V

$\implies$ T₁ â—¦ T₂ (x) = 0_v $\forall$ x$\in$ V

$\implies$ T₁ â—¦ T₂ = 0_v

Now to show that rank(T₁)+rank(T₂) ≤ dim(V)

Note that ker(O_V ) = V for any v ∈ V , O_V(v) = 0, so V ⊆ ker(O_V ); on the other hand, we always have ker(O_V ) ⊆ V . Hence nullity(O_V ) = dim(V).

$\because$ if V be a finite-dimensional vector space. Let S: V → V and T: V → V be linear operators on V. then,

nullity(S ◦ T) ≤ nullity(S) + nullity(T)

using the above concept: we get,

nullity(T₁ â—¦ T₂) ≤ nullity(T₁) + nullity(T₂)

to get,

dim(V ) ≤ nullity(T₁) + nullity(T₂)

Adding dim(V ) to both sides of the inequality and bringing the two terms on the RHS to the LHS, we get,

dim(V ) − nullity(T₁) + dim(V ) − nullity(T₂) ≤ dim(V )

Finally, we apply the rank-nullity theorem twice to get,

rank(T₁)+rank(T₂) ≤ dim(V)