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.
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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($W$) = 2So, 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 mapSo, 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://cheenta.com/collegeprogram/" header_image_url="https://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"]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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