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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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