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## 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

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