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# Kernel of a linear transformation | ISI MStat 2016 Problem 4 | PSB Sample This is a beautiful problem from ISI MStat 2016 Problem 4 PSB (sample) based on Vector space. It uses several concepts to solve it. We provide a detailed solution with prerequisites mentioned explicitly.

## Problem- ISI MStat 2016 Problem 4

For each $c \in \mathbb{R},$ define a function $T_{c}: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}$ by
$T_{c}\left(x_{1}, x_{2}, x_{3}, x_{4}\right):=\left((1+c) x_{1}, x_{2}+c x_{3}, x_{3}+c x_{2},(1+c) x_{4}\right)$
For every $c \in \mathbb{R},$ find the dimension of the null space of $T_{c}$.

## Prerequisites

• kernel or Null space of a linear transformation
• Dimension
• Spanning & Linearly Independent vectors of a vector space

## Solution

Here we have to find the Kernel or null space of $T_{c}$ i.e { $\vec{x}$ : $T_{c} ( \vec{x} )=\vec{0}$ } .

$T_{c}$ is defined as $T_{c}\left(x_{1}, x_{2}, x_{3}, x_{4}\right):=\left((1+c) x_{1}, x_{2}+c x_{3}, x_{3}+c x_{2},(1+c) x_{4}\right)$

So, $T_{c} ( \vec{x} )=\vec{0} \Rightarrow ((1+c) x_{1}, x_{2}+c x_{3}, x_{3}+c x_{2},(1+c) x_{4}) = (0,0,0,0)$ , which gives

(i) $(1+c)x_{1}=0 \Rightarrow x_{1} =0$ if $c \ne -1$

(ii)$(1+c)x_{4}=0 \Rightarrow x_{4} =0$ if $c \ne -1$

(iii) $x_{2}+c x_{3} =0 \Rightarrow x_{2}=-c x_{3}$

(iv) $x_{3}+c x_{2} \Rightarrow x_{3}=-c x_{2}$

(iii) & (iv)$\Rightarrow x_{2}=-c x_{3}=c^2 x_{2} \Rightarrow x_{2} (1-c^2) =0 \Rightarrow x_{2}=0$ if $c \ne \pm 1$

And if $x_{2}=0$ then $x_{3}= 0$ if c $\ne 0$ .

Now for different values of c and using (i),(ii),(iii) and (iv) we will find the Null space $(N(T_{c}) )$ as follows ,

$N(T_{c})$ = $\begin{cases} (x_{1} ,x_{2},x_{2} , x_{4}) & , c=-1 , and { x_{1} ,x_{2}, , x_{4}} \in \mathbb{R} \\ (0,0,0, 0) & c=0 \\ (0,x_{2} ,-x_{2} , 0) &, c=1 , and { x_{2}} \in \mathbb{R} \\ (0,0,0,0) & , c \ne 0,-1,1 \end{cases}$

Therefore for different values of c we will get different dimension of $N(T_{c})$ as follows ,

If c=-1 then $N(T_{c}) = (x_{1} ,x_{2},x_{2} , x_{4} ) = x_{1}(1,0,0,0) + x_{2} (0,1,1,0) + x_{4}(0,0,0,1)$ . Hence the vectors {(1,0,0,0) , (0,1,1,0) ,(0,0,0,1) } spans $N(T_{c})$ and they are Linearly Independent . Thus on this case dimension of null space $N(T_{c})$ is 3 .

If c=0 then $N(T_{c}) =(0,0,0 , 0)$ . Thus on this case dimension of null space $N(T_{c})$ is 0.

If c=1 then $N(T_{c}) =(0,x_{2} ,-x_{2} , 0) = x_{2} (0,1,-1,0)$ .Hence the vectors { (0,1,-1,0) } spans $N(T_{c})$ and they are Linearly Independent . Thus on this case dimension of null space $N(T_{c})$ is 1 .

Finally if $c \ne -1,0,1$ then $N(T_{c}) =(0,0,0,0)$ . Thus on this case dimension of null space $N(T_{c})$ is 0.

This is a beautiful problem from ISI MStat 2016 Problem 4 PSB (sample) based on Vector space. It uses several concepts to solve it. We provide a detailed solution with prerequisites mentioned explicitly.

## Problem- ISI MStat 2016 Problem 4

For each $c \in \mathbb{R},$ define a function $T_{c}: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}$ by
$T_{c}\left(x_{1}, x_{2}, x_{3}, x_{4}\right):=\left((1+c) x_{1}, x_{2}+c x_{3}, x_{3}+c x_{2},(1+c) x_{4}\right)$
For every $c \in \mathbb{R},$ find the dimension of the null space of $T_{c}$.

## Prerequisites

• kernel or Null space of a linear transformation
• Dimension
• Spanning & Linearly Independent vectors of a vector space

## Solution

Here we have to find the Kernel or null space of $T_{c}$ i.e { $\vec{x}$ : $T_{c} ( \vec{x} )=\vec{0}$ } .

$T_{c}$ is defined as $T_{c}\left(x_{1}, x_{2}, x_{3}, x_{4}\right):=\left((1+c) x_{1}, x_{2}+c x_{3}, x_{3}+c x_{2},(1+c) x_{4}\right)$

So, $T_{c} ( \vec{x} )=\vec{0} \Rightarrow ((1+c) x_{1}, x_{2}+c x_{3}, x_{3}+c x_{2},(1+c) x_{4}) = (0,0,0,0)$ , which gives

(i) $(1+c)x_{1}=0 \Rightarrow x_{1} =0$ if $c \ne -1$

(ii)$(1+c)x_{4}=0 \Rightarrow x_{4} =0$ if $c \ne -1$

(iii) $x_{2}+c x_{3} =0 \Rightarrow x_{2}=-c x_{3}$

(iv) $x_{3}+c x_{2} \Rightarrow x_{3}=-c x_{2}$

(iii) & (iv)$\Rightarrow x_{2}=-c x_{3}=c^2 x_{2} \Rightarrow x_{2} (1-c^2) =0 \Rightarrow x_{2}=0$ if $c \ne \pm 1$

And if $x_{2}=0$ then $x_{3}= 0$ if c $\ne 0$ .

Now for different values of c and using (i),(ii),(iii) and (iv) we will find the Null space $(N(T_{c}) )$ as follows ,

$N(T_{c})$ = $\begin{cases} (x_{1} ,x_{2},x_{2} , x_{4}) & , c=-1 , and { x_{1} ,x_{2}, , x_{4}} \in \mathbb{R} \\ (0,0,0, 0) & c=0 \\ (0,x_{2} ,-x_{2} , 0) &, c=1 , and { x_{2}} \in \mathbb{R} \\ (0,0,0,0) & , c \ne 0,-1,1 \end{cases}$

Therefore for different values of c we will get different dimension of $N(T_{c})$ as follows ,

If c=-1 then $N(T_{c}) = (x_{1} ,x_{2},x_{2} , x_{4} ) = x_{1}(1,0,0,0) + x_{2} (0,1,1,0) + x_{4}(0,0,0,1)$ . Hence the vectors {(1,0,0,0) , (0,1,1,0) ,(0,0,0,1) } spans $N(T_{c})$ and they are Linearly Independent . Thus on this case dimension of null space $N(T_{c})$ is 3 .

If c=0 then $N(T_{c}) =(0,0,0 , 0)$ . Thus on this case dimension of null space $N(T_{c})$ is 0.

If c=1 then $N(T_{c}) =(0,x_{2} ,-x_{2} , 0) = x_{2} (0,1,-1,0)$ .Hence the vectors { (0,1,-1,0) } spans $N(T_{c})$ and they are Linearly Independent . Thus on this case dimension of null space $N(T_{c})$ is 1 .

Finally if $c \ne -1,0,1$ then $N(T_{c}) =(0,0,0,0)$ . Thus on this case dimension of null space $N(T_{c})$ is 0.

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