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# Linear Transformation - IIT JAM 2018 Question Number 21

## 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

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]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.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.2.2"]IIT JAM 2018, Question Number 21[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" inline_fonts="Abhaya Libre" open="off"]

### Linear transformation and Vector Space.

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]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. $\bullet \quad\textbf{The correct option is} \quad[\textbf{C}]$[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]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.  [/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]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.  [/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]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[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" custom_margin="20px||20px||false|false" global_module="50750"][et_pb_fullwidth_header title="College Mathematics Program" button_one_text="Learn more" button_one_url="https://www.cheenta.com/collegeprogram/" header_image_url="https://www.cheenta.com/wp-content/uploads/2018/03/College-1.png" _builder_version="4.2.2" background_color="#12876f" custom_button_one="on" button_one_text_color="#12876f" button_one_bg_color="#ffffff"]

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