(adsbygoogle = window.adsbygoogle || []).push({}); Linear Algebra -- Span of vectors

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Linear Algebra - Span of vectors

**Physics Forums | Science Articles, Homework Help, Discussion**