# What is the relationship between spans that contain some of the same vectors?

I have been given the following problem:

Let x, y, z be non-zero vectors and suppose w = 4x + y -3z.

a) If z = 4x + y, then w = _x + _y.

b) Using the calculation in (a), mark the statements below that must be true.

Then I'm given a number of statements of the same type. I will quote only the one for purposes of this discussion and hope to be able to do the rest myself by the end of it!

(i) Span (w, y, z) = Span (w, x)

For reference, I got w = -8x - 2y and this has been marked as correct (whew!)

I'm confused by this question on a couple of fronts. Firstly, I understood a span to be the set of all linear combinations of all vectors in a given space. However, the equation given for w looks like a specific linear combination (i.e. 4 of x and 1 of y etc). Is my understanding of the definition incorrect? Is it actually any linear combination? Am I wrong about w and it's not actually a space (I see the question doesn't say it is) Have I lost the plot completely and this has nothing to do with anything?!!!

I'm also confused about how to engage with this question. Should I be taking the equations for w, y and z and summing them together to see if I get the same as the RHS? (i.e. the sum of w and x?)

I would really appreciate any light you guys can throw on this. I have a ream of questions to answer on spans and this is one of the intro ones. It's clear to me that there's something I'm not getting.

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Ok so the span of a bunch of vectors is intuitively "the vectors that you can make with these ones". By make I mean by using only the operations of addition/subtraction/scalar multiplication. The span makes a vector space.

So in your question the fact that $w = 4x + y - 3z$ is telling you that "$w$ can be made from $x,y,z$". Formally this means $w\in\text{span}\{x,y,z\}$.

But the next fact, that $z = 4x + y$ is telling you that actually "$z$ can be made from $x,y$", i.e. $z\in\text{span}\{x,y\}$.

So really from this you can conclude that "$w$ can be made from $x$ and $y$ only". Indeed, replace $z$ in the definition of $w$ and you get $w = 4x + y - 3(4x+y) = -8x - 2y$, so that $w\in\text{span}\{x,y\}$.

Anyway, for your second point, lets have a think about it. The statement that $\text{span}\{w,y,z\} = \text{span}\{w,x\}$ is saying "can anything made from $w,y,x$ be made only from $w,x$ and conversely, can everything made from $w,x$ be made from $w,y$ and $z$?"

Let's answer the first of these. Well clearly we can make $w$ from $w$ and $x$. Also we have just seen that $w$ can be made from $x$ and $y$, so rearranging, we can make $y$ from $w$ and $x$. Also $z$ is made from $x$ and $y$ and $y$ is made from $w$ and $x$ so that $z$ can be made from $w$ and $x$. Thus anything made from $w,y,z$ CAN be made from $w$ and $x$.

The opposite inclusion is similar. We see that $w$ can obviously be made from $w,y$ and $z$. Also $x$ can be made from $w,y$ and $z$. So anything made from $w$ and $y$ CAN be made from $w,y$ and $z$.

Hope this helps.

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As a geometric picture lets look in $\mathbb{R}^2$. If you take the span of the vector $(1,0)$ you get all vectors of the form $(a,0)$, i.e. the $x$-axis. Similarly for the span of any vector here you get the line defined by it. If you take two vectors you are likely to span the whole space, e.g. span of $(1,0)$ and $(0,1)$ is the whole plane. However if you take two linearly dependent vectors like $(1,1)$ and $(2,2)$ then you just get a line from the span...the problem was that you could "make" $(2,2)$ from $(1,1)$. – fretty Feb 26 '12 at 12:34
Yes, that is very helpful thanks! :) I noticed you looked at the relationship in both directions - will I always need to do that? I was thinking about your geometric picture and how it seems obvious that I could get everything in certain planes from a given span of 3D space, but not vice versa. Is that why you checked both ways? – Katherine Rix Feb 26 '12 at 12:47
Well the span is a set of things...and to show equality of sets you have to go both ways. – fretty Feb 26 '12 at 13:11