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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.

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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 $T : V \longrightarrow V$defined 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
Vector Space
Easy
##### Suggested Book
Abstract Algebra By S.K Mapa

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 $V$is $\{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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