For each problem below, either prove that the mapping is linear or explain why it cannot be linear. 1. f(x_1,x_2 )=(2x_1-x_2,3x_1+x_2) 2. L(x,y,z)=(x+y,y+z,z+5) 3. L(x,y)=(x+y,0,x-2y) 4. f(x,y)=(2x+y,-3x+5y) 5. f(x,y)=(x^2,x+y) 6. L(x,y)=(x,x+y,-y)

jernplate8 2021-01-05 Answered
For each problem below, either prove that the mapping is linear or explain why it cannot be linear.
1.f(x1,x2)=(2x1x2,3x1+x2)
2.L(x,y,z)=(x+y,y+z,z+5)
3.L(x,y)=(x+y,0,x2y)
4.f(x,y)=(2x+y,3x+5y)
5.f(x,y)=(x2,x+y)
6.L(x,y)=(x,x+y,y)
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Expert Answer

lobeflepnoumni
Answered 2021-01-06 Author has 99 answers

Let V and W be vector spaces, and let f:VWf:VW be a function from V to W. (That is, for each vector v\inV, f(v) denotes exactly one vector of W.) Then f is a linear transformation if and only if
f(v1+cv2)=f(v1)+cf(v2),v1,v2V,cR.
Let V and W be vector spaces, and let L:VW be a linear transformation. Let 0V be the zero vector in V and 0W be the zero vector in W. Then L(0V)=0W.​ (1)Heref(x1,x2)=(2x1x2,3x1+x2)f(x).
Then for (x1,x2),(y1,y2)R2(x) f((x1,x2)+c(y1,y2))=f(x1+cy1,x2+cy2)=(2(x1+cy1)(x2+cy2),3(x1+cy1)+(x2+cy2))=(2x1x2+c(2y1y2),3x1+x2+c(3y1+y2))=(2x1x2,3x1+x2)+c(2y1y2,3y1+y2))=f(x1,x2)+cf(y1,y2).
Therefore ff is a linear transformation.
(2)HereL(x,y,z)=(x+y,y+z,z+5)L(x,y,z)=(x+y,y+z,z+5). Since f(0,0,0)=(0,0,5)f(0,0,0)=(0,0,5) not equal to (0,0,0)(0,0,0), therefore LL is not a linear transformation.
(3, 4) Try yourself using the definition.
(5) Wheather f(2,2)=f((1,1)+(1,1)=f(1,1)+f(1,1)?f(2,2)=f((1,1)+(1,1)=f(1,1)+f(1,1)?
(6) Try yourself using the definition.

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