[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/02/IIT-JAM-_2018_21-1.png" alt="calculation of mean and median- AMC 8 2013 Problem" title_text=" mean and median- AMC 8 2013 Problem" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="429px" height="189px" max_height="198px" custom_padding="10px|10px|10px|10px|false|false"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]
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"]
[/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"]
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.
[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/02/IIT-JAM-_2018_21-1.png" alt="calculation of mean and median- AMC 8 2013 Problem" title_text=" mean and median- AMC 8 2013 Problem" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="429px" height="189px" max_height="198px" custom_padding="10px|10px|10px|10px|false|false"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]
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"]
[/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"]
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.
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"]
