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May 18, 2020

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.

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