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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

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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

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