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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.