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College Mathematics

Rank:IIT JAM 2018 PROBLEM 9

This problem appeared in IIT JAM 2018. This problem appeared from the brach of vector space which also requied basic knowledge from rank of a matrix.

Let’s Try A Warm Up MCQ

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Understand the problem

Consider the vector space V over \mathbb{R}of the polynomial functions of degree less than or equal to 3 defined on \mathbb{R}. Let T : V \longrightarrow Vdefined by $latex (Tf)(x) = f(x)-xf'(x). Then the rank of T is  (a) 1  (b) 2 (c) 3 (d) 4 
Source of the problem
IIT JAM 2018 Problem 9
Topic
Vector Space 
Difficulty Level
Easy
Suggested Book
Abstract Algebra By S.K Mapa

Start with hints

Do you really need a hint? Try it first!
Rank(T) = dim(Range(T)) There is one easy way to calculate rank of every linear transformation. Step 1:  Take by basis  \beta= \{e_1,....,e_n\}of the vector space V. Step 2: Write down the matrix [T]_{\beta}^{\beta}Step 3: Calculate the rank of the matrix  [T]_{\beta}^{\beta}Now can you follow these steps to get the answer?  
Standard Basis of Vis \{1,x,x^{2},x^{3}\} = \beta(Tf) (x) =f(x) - xf^{'}(x)(T1) (x) = 1 - 0 = 1; (Tx) (x) = x - x = 0; (T x^{2}) (x)= x^{2} - 2x^{2} = -x^{2}; (T x^{3}) (x) = -2x^{3}So, [T]_{\beta}^{\beta} = \begin{pmatrix}  1 & 0 & 0 & 0 \\  0 & 0 & 0 & 0\\  0 & 0 & -1 & 0\\  0 & 0 & 0 & -2 \\  \end{pmatrix}Hence the rank is 3

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