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HallsofIvy
#3
Apr21-11, 09:01 PM
Math
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Saying there is a unique linear tranformation satisfying certain conditions, means there is one and only one such transformation.

For example, if we are only told that the linear transformation L maps (1, 0), in R2 to (1, 1), then we know that L(x,y)= L(x(1, 0)+ y(0, 1))= xL(1,0)+ yL(0,1)= x(1,1)+ yL(0,1)= (x,x)+ yL(0,1). But I don't know what L(0,1) is. Since (0,1) and (1, 0) are independent, I cannot get L(0,1) from "L(1, 0)= (1,1)". There are many such (actually an infinite number) of Linear transformation satisfying that condition.

In general, a linear transformation on an n-dimensional vector space is completely determined by its action on n independent vectors (which then must be a basis for the space).

But if I am told that L maps (1, 0) to (1, 1) and maps (0, 1) to (2, 3), then I can say that L(x,y)= L(x(1,0)+ y(0,1))= xL(1,0)+ yL(0,1)=x(1,1)+ y(2,3)= (x+2y, x+ 3y). That is the only limear transformation satisfying those conditions- it is unique.

In the example, T(9x+5) = (.1,.2) and T(7x+4) = (.3,.8), "9x+ 5" and "7x+ 4" are independent and, since the set of linear polynomials has dimension 2, form a basis for that space. Given any such polynomial, ax+ b, it is easy to show that ax+ b= (4a- 7b)(9x+ 5)+ (9b-5a)(7x+ 4) so that T(ax+ b)= (4a- 7b)T(9x+5)+ (9b-5a)T(7x+4)= (4a+ 7b)(.1,.2)+ (9b-5a)(.3, .8)= (-1.1a+ 3.4b, -3.2a+ 8.6), a unique linear transformation.