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Span question with Restrictions

  • Thread starter jberg074
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
22,097
3,277
Hi jberg074! :smile:

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

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

and you'll find your conditions on a,b,c and d.
 
  • #3
8
0
Is the answer as simple as this?
a=6b-c
b=b
c=2b-a/2
d=4b+a/2
 
  • #4
22,097
3,277
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

[tex]\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}[/tex]
 
  • #5
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[tex]\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}[/tex][/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"?
 
  • #6
22,097
3,277
[tex]\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}[/tex]
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

[tex](a,b,c,d)=(a,b,2b-a/2,4b+a/2)[/tex]
 
  • #7
8
0
Ah, I understand now. I wasn't exactly sure what the question was asking for at first; thank you very much for the help!
 

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