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Linear Algebra -- Span of vectors
Let x, y, z be non-zero vectors and suppose w=15x−10y−3z. z=3x−2y, then w=?
Using the calculation above, mark the statements below that are true.
A. Span(w,x) = Span(w,x,y)
B. Span(w,z) = Span(y,z)
C. Span(x,y,z) = Span(w,y)
D. Span(w,y,z) = Span(w,z)
E. Span(w,y,z) = Span(x,y)
w=6x - 4y
Since w and z are multiples of each other, their span would be equal. Span(w)=Span(z)
I'm not sure if my understanding of span of vectors are entirely correct, but this is my attempt at the first statement (A):
span(w, x) = { c1*w + c2*x
= c1*(6x - 4y) + c2*y
= c1*6x - c1*4y + c2*y = B
where c1 and c2 are real numbers and B is some vector}
Span(w,x,y) = { c1*w + c2*x + c3*y
= c1*(6x - 4y) + c2*x + c3*y
= c1*6x - c1*4y + c2*x + c3*y = B
where c1, c2, and c3 are real numbers and B is some vector}
Since there are essentially only two vectors in both spans (x vector and y vector), both spans are equivalent. For span(w, x), all multiples of vector x can be chosen through c1 and all multiples of vector y can be chosen through c2. For span(w, x, y), all multiples of vector x can be chosen through c2 and all multiples of vector y can be chosen through c3 even if c1=0.
So statement A is true.
Am I approaching this problem the wrong way? If I continue this solving strategy then all spans are expressed in term of vectors x and y so they are all true statements, but I don't think that is the answer. Any input is greatly appreciated.
Homework Statement
Let x, y, z be non-zero vectors and suppose w=15x−10y−3z. z=3x−2y, then w=?
Using the calculation above, mark the statements below that are true.
A. Span(w,x) = Span(w,x,y)
B. Span(w,z) = Span(y,z)
C. Span(x,y,z) = Span(w,y)
D. Span(w,y,z) = Span(w,z)
E. Span(w,y,z) = Span(x,y)
Homework Equations
The Attempt at a Solution
w=6x - 4y
Since w and z are multiples of each other, their span would be equal. Span(w)=Span(z)
I'm not sure if my understanding of span of vectors are entirely correct, but this is my attempt at the first statement (A):
span(w, x) = { c1*w + c2*x
= c1*(6x - 4y) + c2*y
= c1*6x - c1*4y + c2*y = B
where c1 and c2 are real numbers and B is some vector}
Span(w,x,y) = { c1*w + c2*x + c3*y
= c1*(6x - 4y) + c2*x + c3*y
= c1*6x - c1*4y + c2*x + c3*y = B
where c1, c2, and c3 are real numbers and B is some vector}
Since there are essentially only two vectors in both spans (x vector and y vector), both spans are equivalent. For span(w, x), all multiples of vector x can be chosen through c1 and all multiples of vector y can be chosen through c2. For span(w, x, y), all multiples of vector x can be chosen through c2 and all multiples of vector y can be chosen through c3 even if c1=0.
So statement A is true.
Am I approaching this problem the wrong way? If I continue this solving strategy then all spans are expressed in term of vectors x and y so they are all true statements, but I don't think that is the answer. Any input is greatly appreciated.