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ISI MStat PSB 2014 Problem 1 | Vector Space & Linear Transformation

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 !!

Problem- ISI MStat PSB 2014 Problem 1


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.

Prerequisites


Linear Transformation

Null Space

Dimension

Solution :

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}.


Food For Thought

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


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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 !!

Problem- ISI MStat PSB 2014 Problem 1


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.

Prerequisites


Linear Transformation

Null Space

Dimension

Solution :

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}.


Food For Thought

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


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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