Question #160247

Let T: R^3→R^2 be given by:

T(x2,x2,x3)= (X1+x2+x3,x2+x3)

Prove that T is a linear transformation. Also find the rank and nullity of T.


1
Expert's answer
2021-02-07T15:11:57-0500

T: R3→R2 be given by:

T(x1,x2,x3)=(x1+x2+x3,x2+x3)

    \implies First, we need to check T is a linear transformation.

let (x1,x2,x3), (y1,y2,y3) \in R3 and α\alpha \isin R

T(α\alpha (x1,x2,x3) + (y1,y2,y3)) = T((α\alphax1,α\alphax2,α\alphax3) + (y1,y2,y3))

= T( α\alphax1+y1, α\alphax2+y2, α\alphax3+y3)

= (α\alphax1+y1+α\alphax2+y2+α\alphax3+y3 , α\alphax2+y2+α\alphax3+y3)

= (α\alpha(x1+x2+x3) + (y1+y2+y3), α\alpha(x2+x3)+(y2+y3))

= α\alpha(x1+x2+x3,x2+x3) + (y1+y2+y3,y2+y3)

= α\alphaT(x1,x2,x3) + T(y1,y2,y3)

since,it is satisfying the definition of linear transformation.

Hence,

T is a linear transformation.


    \implies Now we need to find the rank and nullity of T.

as given;

T(x1,x2,x3)=(x1+x2+x3,x2+x3)

here first to find the matrix corresponding to this linear transformation,

let β\beta ={(1,0,0),(0,1,0),(0,0,1)}

T(1,0,0) = (1,0) = 1(1,0) + 0(0,1)

T(0,1,0) = (1,1) = 1(1,0) + 1(0,1)

T(0,0,1) = (1,1) = 1(1,0) + 1(0,1)


[T] =[111011]\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}


since matrix have two non-zero rows. therefore rank of a matrix is 2.

    \implies rank (T)= 2

using rank nullity theorem which states that rank(T) + nullity(T) = n;

rank(T) + nulity(T) = 3 \because n = 3

nulity(T) = 3 - rank(T) = 3 - 2 =1

    \implies nulity(T) = 1



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