Ranking Rank of an operator: TIFR 2019 GS Part A, Problem 5

Understand the problem

Let \(V\) be an \(n\)-dimensional vector space and let \(T : V \to V\) be a linear transformation such that \(Rank T  \leq Rank T^3\).

Let \(V\) be an \(n\)-dimensional vector space and let \(T : V \to V\) be a linear transformation such that \(Rank T  \leq Rank T^3\).

Then which one of the following statements is necessarily true?

  • \(Null space( T ) = Range(T )\)
  •  \(Null space( T )\cap Range(T ) = {0}\)
  • There exists a nonzero subspace \(W\) of \(V\) such that \(Null space(T )\cap Range(T ) = W\)
  • \(Null space( T )\subseteq Range(T )\)

Start with hints

Source of the problem
TIFR 2019 GS Part A, Problem 4
Topic
Linear algebra
Difficulty Level
Easy
Suggested Book
Introduction to Linear Algebra by Gilbert Strang
Do you really need a hint? Try it first!

\(Rank(T) \leq Rank(T^3)\) so, \(Null(T^3)\leq Null (T)\)[By Rank Nullity theorem]
\(Null(T)\subseteq Null (T^3)\), hence \(Null(T)= Null (T^3)\)

Let \(x\in N(T) \cap R(T)\), \(Tx=0\); claim: \(x=0\).

Let \(\exists y\in V\) s.t \(Ty=x \) hence \(T^2(Ty)=T^2(x)=0\)

hence \(T^3(y)=0\)

hence, \(y \in N(T^3)=N(T)\)

Hence, \(x=Ty\).

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