Span question with Restrictions

[jberg074]

Homework Statement

Let U be the span of
[ 0 ] [ 2 ]
[ 1 ] and [ 0 ]
[ 2 ] [ -1]
[ 4 ] [ 1 ]

Give conditions on a, b, c and d so that [a b c d] (transposed) is in U (as in, give restrictions that are equations of only a, b, c and d.

Homework Equations

The Attempt at a Solution

I've set up the problem to set the two given vectors multiplied by scalar x and y, yielding four equations:
2y = a,
x = b,
2x - y = c, and
4x + y = d.

Is this the right first step to take? After that, I take x or b as a free variable, and try to rewrite the other equations in terms of b, yielding:

2y = a, b = b, 2b - y = c, and 4b + y = d.

After this, however, I am stuck. Any pointers? Did I set up the problem the right way?

Edit: I also know that [a b c d] must be a linear combination of the two given vectors if it is in the span. However, if it is a linearly independent set, the trivial zero vector could be matched with [a b c d], although I am pretty sure the answer isn't that simple.

 

[micromass]

Hi jberg074!

You also know that 2y=a, so y=\frac{a}{2}. Substitute that into the equations

2b - y = c
4b + y = d.

and you'll find your conditions on a,b,c and d.

 

[jberg074]

Is the answer as simple as this?
a=6b-c
b=b
c=2b-a/2
d=4b+a/2

 

[micromass]

Is the answer as simple as this?
a=6b-c
b=b
c=2b-a/2
d=4b+a/2

I'm not sure where you got the a=6b-c, actually. But the two last equations are what you need. These give necessary and sufficient conditions for something to be in the span. In general, the span is given by

\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}

 

[jberg074]

\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}[/QUOTE]

I used other questions to eliminate variables and to try and get a in terms of the other. However, in the text above, shouldn't "a" be replaced by "a/2"?

 

[micromass]

\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}

I used other questions to eliminate variables and to try and get a in terms of the other. However, in the text above, shouldn't "a" be replaced by "a/2"?[/QUOTE]

No, we have concluded that

a=a
b=b
c=2b-a/2
d=4b+a/2

So, we will have

(a,b,c,d)=(a,b,2b-a/2,4b+a/2)

 

[jberg074]

Ah, I understand now. I wasn't exactly sure what the question was asking for at first; thank you very much for the help!