## 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:-

## 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

##### Suggested Book

## 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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