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.
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 \(T_{c}\).
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.
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 \(T_{c}\).
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.
