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# 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 $$Null space(T )\cap Range(T ) = W$$
• $$Null space( T )\subseteq Range(T )$$

##### Source of the problem
TIFR 2019 GS Part A, Problem 4
Linear algebra
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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