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Linear Transformation – IIT JAM 2018 Question Number 21

Try this beautiful problem from IIT JAM 2018. It involves the concept of vector spaces and linear mapping. We provide sequential hints so that you can try the problem.

What are we learning ?

Competency in Focus: Linear Transformation

This problem from IIT JAM 2018 is based on calculation of linear transformation. It is Question no. 21 of the IIT JAM 2018 Problem series.

First look at the knowledge graph:-

calculation of  mean and median- AMC 8 2013 Problem

Next understand the problem

Let $U, V, \textbf{and}\quad W$ be finite dimensional real vector spaces, $T:U\rightarrow V, S: V\rightarrow W, P: W\rightarrow U$ be linear transformations. If range($ST$)=nullspace($P$), nullspace($ST$)=range ($P$) and rank($T$)=rank($S$), Then which one of the following is true? $\textbf{(A)}$ nullity of $T$ = nullity of $S$ $\textbf{(B)}$ dimension of $U \neq $  dimension of $W$ $\textbf{(C)}$ If dimension of $V = 3$, If dimension of $U = 4$, then $P$ is not identically zero. $\textbf{(C)}$ If dimension of $V = 4$, If dimension of $U = 3$, and $T$ is one-one then $P$ is identically zero. 
Source of the problem
IIT JAM 2018, Question Number 21
Key Competency

Linear transformation and Vector Space.

Difficulty Level
7/10

Start with hints 

Do you really need a hint? Try it first!
I want to make an opinion that this question is not hard but it is time consuming. So in a time paced exam beware of this kind of question. The fact I’ve used here : i) Ker$(ST)={0} \iff ST \quad \textbf{is injective}$ ii) $ST\quad \textbf{is injective} \iff  $T$ \quad \textbf{is injective}$ iii) And creating the counter examples which are  most time consuming. So, in hint 1, I want to disclose the answer and I can first try to find out the counter example by yourself. $\bullet \quad\textbf{The correct option is} \quad[\textbf{C}]$
Consider $U=V=W=\mathbb{R}^2$ and the maps are  The range($ST$) =$S_p\{c_1\}=$ Nullspace($P$). range($P$)=$S_p\{c_2\}=$ Nullspace($ST$) & rank($T$) $=1=$ rank ($S$) But, Nullity of $T=1 \neq 2=$ Nullity of $S$. And dim($U$) = dim($W$) = 2 So, the option $(A)$ and $(B)$ are incorrect.  
Consider $U= \mathbb{R}^3,V= \mathbb{R}^4, W= \mathbb{R}^5 $ range ($ST$) = $S_p\{c_1,c_2\}=$ Nullspace($P$) Nullspace ($ST$)=$S_p\{c_3\}$=Range($P$) rank($T$)=$2$=rank($S$) and $P$ is not zero map So, option $(D)$ is not correct. Hence option $(C)$ is the one left which has to be true, Now lets prove that.  
We’ll prove that $P$ is non zero. Suppose $P=0$ $U\longrightarrow V\longrightarrow W$ dim($U$) = $4$, dim($V$) =$3$  Now,         range ($P$)  $=0=$ Nullspace ($ST$)         $\Rightarrow ST $ is injective         $\Rightarrow T  $ is injective         $\Rightarrow  $ dim ($V\geq 4$)         Which is a contradiction. Hence $P=0$ And we are done

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