Linear algebra Q: show spans of vectors are equal

[SophieToft]

Hi

if two (a,b,c) and (d,e,f) can be written as linear combinations for two other vectors (x,y,z) and (t,u,p)

thereby meaning

ax1 + dx2 = x
bx1 +ex2 =y
cx1 + fx2 = z

and

ax1 + dx2 = t
bx1 +ex2 =u
cx1 + fx2 = p

how do I then show from this that span((a,b,c),(d,e,f)) = span((x,y,z),(t,u,p))

Can I claim here that since the set of vector in R^3 (a,b,c),(d,e,f) are linear combinations of both (x,y,z),(t,u,p) respectively, and since R^3 is sub-plane or R^n then there respective spans will always equal each other?

Or this there a better explaination for this?

Sincerely Yours
Sophie Toft

 

[JakeD]

Hi

if two (a,b,c) and (d,e,f) can be written as linear combinations for two other vectors (x,y,z) and (t,u,p)

thereby meaning

ax1 + dx2 = x
bx1 +ex2 =y
cx1 + fx2 = z

and

ax1 + dx2 = t
bx1 +ex2 =u
cx1 + fx2 = p

how do I then show from this that span((a,b,c),(d,e,f)) = span((x,y,z),(t,u,p))

Can I claim here that since the set of vector in R^3 (a,b,c),(d,e,f) are linear combinations of both (x,y,z),(t,u,p) respectively, and since R^3 is sub-plane or R^n then there respective spans will always equal each other?

Or this there a better explaination for this?

Sincerely Yours
Sophie Toft

Hi. It may not be true that span((a,b,c),(d,e,f)) = span((x,y,z),(t,u,p)). Suppose the vectors (x,y,z) and (t,u,p) are independent but (a,b,c) = (d,e,f). Then those two vectors are both linear combinations but the spans are not equal. An additional hypothesis is needed, and with it the proof may be more obvious.

 

[SophieToft]

Hello,

Would You please elaborate?

Do You mean I need to conclude that the vectors are linear independent ?

Sincerley

Sophie Toft

Hi. It may not be true that span((a,b,c),(d,e,f)) = span((x,y,z),(t,u,p)). Suppose the vectors (x,y,z) and (t,u,p) are independent but (a,b,c) = (d,e,f). Then those two vectors are both linear combinations but the spans are not equal. An additional hypothesis is needed, and with it the proof may be more obvious.[/quote]

 

[JakeD]

This is a response to your second post, but I'll do that by showing some problems with your first post.

Hi

if two (a,b,c) and (d,e,f) can be written as linear combinations for two other vectors (x,y,z) and (t,u,p)
linear combinations "for"? Shouldn't that be "of"?

thereby meaning

ax1 + dx2 = x
bx1 +ex2 =y
cx1 + fx2 = z

and

ax1 + dx2 = t
bx1 +ex2 =u
cx1 + fx2 = p

These equations say (x,y,z) and (t,u,p) are linear combinations of (a,b,c) and (d,e,f), not the other way around as you start off with.

how do I then show from this that span((a,b,c),(d,e,f)) = span((x,y,z),(t,u,p))

Using the equations you have written, suppose x1 and x2 are the same in both. Then (x,y,z) = (t,u,p) and their span has dimension 1. If (a,b,c) and (d,e,f) are independent, their span has dimension 2 and the two spans are not equal. So in this example, some additional condition is needed to guarantee x1 and x2 are not the same (or the same multiples of each other) in the two sets of equations.

Can I claim here that since the set of vector in R^3 (a,b,c),(d,e,f) are linear combinations of both (x,y,z),(t,u,p) respectively, and since R^3 is sub-plane or R^n then there respective spans will always equal each other?

Or this there a better explaination for this?

Sincerely Yours
Sophie Toft

I suggest that you state the original problem you were given and then your interpretation of it.

 

[SophieToft]

very well here it comes.

(a)

show that the vector (2,7,6) be we written as a linear combination of the vectors

(1,3,2) and (0,1,2)

(b) show that the vector (-1,0,4) can be written as a linear combination of the vectors (1,3,2) and (0,1,2)

(c)

show that Span((1,3,2),(0,1,2)) = Span( (2,7,6), (-1,0,4))

My solution (a).

I write vectors in equation form

\(\displaystyle x_1 \[ \begin{array}{c} 1 \\ 3 \\ 2 \end{array} \] + x_2 \[ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \] = \[ \begin{array}{c} 2 \\ 7 \\ 6 \end{array} \]\)

which can be rewritten to

\(\displaystyle \[ \begin{array}{c} x_{1} \\ 3x_{1} + x_{2} \\ 2x_{1} + 2x_{2} \end{array} \] = \[ \begin{array}{c} 2 \\ 7 \\ 6 \end{array} \]\)

There must exist x_1 and x_2 which makes the above set of equations true.

These are found be rewritten the system into its equivalent coefficient matrix.

\(\displaystyle \[ \begin{array}{cc} 1 & 0 & 2 & \\ 3 & 1 & 7 \\ 2 & 2 & 6 \end{array} \] \mathrm\)

using row reduction I get

\(\displaystyle \[ \begin{array}{cc} 1 & 0 & 2 & \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \] \mathrm\)

which gives x_1 = 2 and x_2 = 1

If I insert into the original equation, then values of x_1 and x_2 make it true.

(b)

Following the same method used in (a) I get x1 = -1 and x_2 = 3.

(c)

How do I use these results to prove that Span((1,3,2),(0,1,2)) = Span((2,7,6),(-1,0,4)) ???

Sincerely Yours
Sophie Toft

 

[pka]

It is perfectly clear that:
\(\displaystyle 2\left( {1,3,2} \right) + \left( {0,1,2} \right) = \left( {2,7,6} \right)\quad \& \quad - \left( {1,3,2} \right) + 3\left( {0,1,2} \right) = \left( { - 1,0,4}. \right)\)

Any linear combination of (2,7,6) & (-1,0,4) can be written as:
\(\displaystyle t\left( {2,7,6} \right) + s\left( { - 1,0,4} \right) = \left( {2t - s} \right)\left( {1,3,2} \right) + \left( {t + 3s} \right)\left( { - 1,0,4} \right).\)

By letting t=1 and s=0 we get (2,7,6) as linear combination of (1,3,2) & (0,1,2).
Do about the same for (-1,0,4). There you have it.

 

[SophieToft]

It is perfectly clear that:
\(\displaystyle 2\left( {1,3,2} \right) + \left( {0,1,2} \right) = \left( {2,7,6} \right)\quad \& \quad - \left( {1,3,2} \right) + 3\left( {0,1,2} \right) = \left( { - 1,0,4}. \right)\)

Any linear combination of (2,7,6) & (-1,0,4) can be written as:
\(\displaystyle t\left( {2,7,6} \right) + s\left( { - 1,0,4} \right) = \left( {2t - s} \right)\left( {1,3,2} \right) + \left( {t + 3s} \right)\left( { - 1,0,4} \right).\)

By letting t=1 and s=0 we get (2,7,6) as linear combination of (1,3,2) & (0,1,2).
Do about the same for (-1,0,4). There you have it.

Okay,

Am I right if I say ?

The conclusion is then that The two vector set will always equals their respective spans because they can we written as linear combination of each other?

Sincerely Sophie