Thread: Linear Algebra Problem Involving Span

I've attached the problem below.

The question asks me to find vectors u and v (in R^3) such that u+v is equal to a vector w with the form

|2b+4c |
| b |
| c |


In my approach, I first evaluated the span{u,v}.
If u=|u1 u2 u3|^T and v=|v1 v2 v3|^T. I don't understand why they are asking me for a specific vector u and v. Couldn't I just let either u2 and u3 or v2 and v3 be zero and u2 and u3 or v2 and v3 not picked be 1?

Since any vector in $\displaystyle w \in W$ can be written as $\displaystyle w = (2b+4c, b, c)$ we can write $\displaystyle w$ as $\displaystyle w = b(2, 1, 0) + c(4, 0, 1)$ so that $\displaystyle w$ is a linear combination of $\displaystyle (2, 1, 0)$ and $\displaystyle (4, 0, 1)$ this means that $\displaystyle W = span\{(2, 1, 0), (4, 0, 1)\}$

Thank you for the reply. I can see how that can work. In general, can't I represent W as the spans of other vectors u and v as long as the components add up to create w with the form specified?

W = span\{(2,1,1), (4,0,0)} What about the span of these two vectors?
What about W = span\{(2,0,0), (4,1,1)}?

Not quite. Consider your example:
$\displaystyle W = span\{(2, 0, 0), (4, 1, 1)}$
This means that for any $\displaystyle w \in W$ we have $\displaystyle w = b(2, 0, 0) + c(4, 1, 1)$ for some scalars $\displaystyle b, c$
this means that $\displaystyle w = (2b + 4c, c, c)$ which is not the same as the original $\displaystyle w = (2b + 4c, b , c)$
In other words, you can't just rearrange those two vectors.


In general if you have a problem like this you want to represent the general case as a linear combinations of some vectors (thus the span). What you would do is separate the coefficients from each other (for example in your original question: $\displaystyle w = (2b, b, 0) + (4c, 0, c)$) then factor out the coefficients (again in your original question $\displaystyle w = b(2, 1, 0) + c(4, 0, 1)$) to obtain a linear combination of some vectors. The span of those vectors will be equal to W. You can also use any scalar multiple of those vectors (for example $\displaystyle W = span\{(2, 1, 0), (4, 0, 1)\} = span\{(4, 2, 0), (4, 0, 1)\} = span\{(2, 1, 0), (-4, 0, -1)\} $ and you can keep going with any scalar multiple).

Does that make sense?

Yes it does. We're essentially finding the span of u and v that makes up the column vectors (2,1,0) and (4,0,1) of a coefficient matrix that is associated with a transformation whose image is of the form {(2b+4c),(b,c)}. In other words, the image of our transformation which has the general result {(2b+4c),(b,c)}, has span(u,v).