L(x,y)=(x-y,y-3x,2y,2x-8y,3x-27y) Where L: R^2--->R^4.
I know that the method involves reducing a matrix and and then using that reduced matrix to put the solution in vector parametric form which produces linearly independent vectors but I'm unsure how to construct the matrix?
Answers will be greatly appreciated :)
I know that the method involves reducing a matrix and and then using that reduced matrix to put the solution in vector parametric form which produces linearly independent vectors but I'm unsure how to construct the matrix?
Answers will be greatly appreciated :)
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Let a be in the image of L, which is in R^5 (not R^4). Then
a = (x-y , y-3x , 2y, 2x-8y , 3x-27y ) for some x and y in R. Thus,
a = x(1, -3, 0, 2, 3) + y(-1, 1, 2, -8, -27),
and so a is a linear combination of the vectors r=(1, -3, 0, 2, 3) and s=(-1, 1, 2, -8, -27).
Conversely, if a vector b is in the span of r and s, then there exists a vector in R^2 such that its image under L is b. Thus the image of L is precisely the subspace spanned by r and s.
a = (x-y , y-3x , 2y, 2x-8y , 3x-27y ) for some x and y in R. Thus,
a = x(1, -3, 0, 2, 3) + y(-1, 1, 2, -8, -27),
and so a is a linear combination of the vectors r=(1, -3, 0, 2, 3) and s=(-1, 1, 2, -8, -27).
Conversely, if a vector b is in the span of r and s, then there exists a vector in R^2 such that its image under L is b. Thus the image of L is precisely the subspace spanned by r and s.