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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.

## 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.

## 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

### Linear transformation and Vector Space.

7/10
##### Suggested Book

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

# College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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