This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 1 based on Vector space and Eigen values and Eigen vectors . Let's give it a try !!
Let \(E={1,2, \ldots, n},\) where n is an odd positive integer. Let \( V\) be
the vector space of all functions from E to \(\mathbb{R}^{3}\), where the vector space
operations are given by \( (f+g)(k) =f(k)+g(k)\), for \( f, g \in V, k \in E \
(\lambda f)(k) =\lambda f(k),\) for \( f \in V, \lambda \in \mathbb{R}, k \in E \)
(a) Find the dimension of \(V\)
(b) Let \(T: V \rightarrow V\) be the map given by \( T f(k)=\frac{1}{2}(f(k)+f(n+1-k)), \quad k \in E \)
Show that T is linear.
(c) Find the dimension of the null space of T.
Linear Transformation
Null Space
Dimension
While doing this problem we will use a standard notation for vectors of canonical basis i..e \( e_j\) . In \(R^{3} \) they are \( e_1=(1,0,0) , e_2=(0,1,0) \) and \( e_3=(0,0,1) \) .
(a) For \( i \in {1 , 2 , \cdots , n}\) and \( j \in {1 , 2 , 3}\) , let \( f_{ij}\) be the function in \(V\) which maps \( i \mapsto e_j\) and \(k \mapsto (0,0,0)\) where \(k \in {1 , 2 , \cdots , n}\) and \( k \neq i\). Then \( {f_{ij} : i \in {1 , 2 , \cdots , n} , j \in {1 , 2 , 3}}\) is a basis of \(V\) .
It looks somewhat like this , \(f_{11}(1)={(1,0,0)} ,f_{11}(2)={(0,0,0)} , \cdots , f_{11}(n)={(0,0,0)} \)
\( f_{12}(1)={(0,1,0)} ,f_{12}(2)={(0,0,0)} , \cdots , f_{12}(n)={(0,0,0)} \) , \( \cdots , f_{n3}(1)={(0,0,0)} ,f_{n3}(2)={(0,0,0)} , \cdots , f_{n3}(n)={(0,0,1)} \)
Hence , dimension of \(V\) is 3n.
(b) To show T is linear we have to show that \( T(af(k)+bg(k)) =aT(f(k))+bT(g(k)) \) for some scalar a,b .
\(T(af(k)+bg(k))=\frac{ af(k)+bg(k)+af(n+1-k)+bg(n+1-k)}{2} = a \frac{f(k)+f(n+1-k)}{2} + b \frac{g(k)+g(n+1-k)}{2} = aT(f(k))+bT(g(k)) \).
Hence proved .
(c) \( f\in ker T\) gives \(f(k)=-f(n+1-k)\) so, the values of \(f\) for the last \(\frac{n-1}{2}\) points are opposite to first \(\frac{n-1}{2}(i.e. f(n)=-f(1) ~\text{etc.})\) so we can freely assign the values of f for first \(\frac{n-1}{2}\) to any of \(e_j\) .Hence, the null space has dimension \(\frac{3(n-1)}{2}.\)
let \( T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be a non singular linear transformation.Prove that there exists a line passing through the origin that is being mapped to itself.
Prerequisites : eigen values & vectors and Polynomials
This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 1 based on Vector space and Eigen values and Eigen vectors . Let's give it a try !!
Let \(E={1,2, \ldots, n},\) where n is an odd positive integer. Let \( V\) be
the vector space of all functions from E to \(\mathbb{R}^{3}\), where the vector space
operations are given by \( (f+g)(k) =f(k)+g(k)\), for \( f, g \in V, k \in E \
(\lambda f)(k) =\lambda f(k),\) for \( f \in V, \lambda \in \mathbb{R}, k \in E \)
(a) Find the dimension of \(V\)
(b) Let \(T: V \rightarrow V\) be the map given by \( T f(k)=\frac{1}{2}(f(k)+f(n+1-k)), \quad k \in E \)
Show that T is linear.
(c) Find the dimension of the null space of T.
Linear Transformation
Null Space
Dimension
While doing this problem we will use a standard notation for vectors of canonical basis i..e \( e_j\) . In \(R^{3} \) they are \( e_1=(1,0,0) , e_2=(0,1,0) \) and \( e_3=(0,0,1) \) .
(a) For \( i \in {1 , 2 , \cdots , n}\) and \( j \in {1 , 2 , 3}\) , let \( f_{ij}\) be the function in \(V\) which maps \( i \mapsto e_j\) and \(k \mapsto (0,0,0)\) where \(k \in {1 , 2 , \cdots , n}\) and \( k \neq i\). Then \( {f_{ij} : i \in {1 , 2 , \cdots , n} , j \in {1 , 2 , 3}}\) is a basis of \(V\) .
It looks somewhat like this , \(f_{11}(1)={(1,0,0)} ,f_{11}(2)={(0,0,0)} , \cdots , f_{11}(n)={(0,0,0)} \)
\( f_{12}(1)={(0,1,0)} ,f_{12}(2)={(0,0,0)} , \cdots , f_{12}(n)={(0,0,0)} \) , \( \cdots , f_{n3}(1)={(0,0,0)} ,f_{n3}(2)={(0,0,0)} , \cdots , f_{n3}(n)={(0,0,1)} \)
Hence , dimension of \(V\) is 3n.
(b) To show T is linear we have to show that \( T(af(k)+bg(k)) =aT(f(k))+bT(g(k)) \) for some scalar a,b .
\(T(af(k)+bg(k))=\frac{ af(k)+bg(k)+af(n+1-k)+bg(n+1-k)}{2} = a \frac{f(k)+f(n+1-k)}{2} + b \frac{g(k)+g(n+1-k)}{2} = aT(f(k))+bT(g(k)) \).
Hence proved .
(c) \( f\in ker T\) gives \(f(k)=-f(n+1-k)\) so, the values of \(f\) for the last \(\frac{n-1}{2}\) points are opposite to first \(\frac{n-1}{2}(i.e. f(n)=-f(1) ~\text{etc.})\) so we can freely assign the values of f for first \(\frac{n-1}{2}\) to any of \(e_j\) .Hence, the null space has dimension \(\frac{3(n-1)}{2}.\)
let \( T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be a non singular linear transformation.Prove that there exists a line passing through the origin that is being mapped to itself.
Prerequisites : eigen values & vectors and Polynomials
