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Math Help - Subspaces and Spanning

  1. #1
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    Subspaces and Spanning

    I'm having some trouble with these 2 question on spanning, my prof has not given us any examples and I can't find any help from the book, any help would be awesome.

    1. Let u,v be two vectors in R^n. Show that

    span{u,v} = span{2u + 3v, u + 2v}

    2. Let u,v,w be two vectors in R^n. Show that

    span{u,v,w} = span{u + v, u + w, v + w}
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  2. #2
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    Quote Originally Posted by bloc182 View Post
    1. Let u,v be two vectors in R^n. Show that

    span{u,v} = span{2u + 3v, u + 2v}
    Show that if \bold{x}\in \text{spam}(\bold{u},\bold{v}) then \bold{x}\ \in \text{spam}(2\bold{u} + 3\bold{v},\bold{u}+2\bold{v}) and that if \bold{x}\ \in \text{spam}(2\bold{u} + 3\bold{v},\bold{u}+2\bold{v}) then \bold{x}\in \text{spam}(\bold{u},\bold{v}).
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  3. #3
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    This is a coordination transformation issue.

    <br />
\mathfrak{A} = (u,v),\mathfrak{B}=(2u+3v,u+2v)<br />
,A=BS,S=\begin{bmatrix} 2&-1\\-3&2  \end{bmatrix},SS^{-1}=I_n<br />
    <br />
\therefore \vec{x} =span(u,v)= A[\vec{x}]_{\mathfrak{A}}=BS[\vec{x}]_{\mathfrak{B}}=span(2u+3v,u+2v)<br />
    Last edited by math2009; March 29th 2009 at 05:03 PM.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Show that if \bold{x}\in \text{spam}(\bold{u},\bold{v}) then \bold{x}\ \in \text{spam}(2\bold{u} + 3\bold{v},\bold{u}+2\bold{v}) and that if \bold{x}\ \in \text{spam}(2\bold{u} + 3\bold{v},\bold{u}+2\bold{v}) then \bold{x}\in \text{spam}(\bold{u},\bold{v}).
    "span", not "spam"!

    Yet another way to show this: Any vector in the span of {2u+3v, u+2v} must be of the form a(2u+3v)+ b(u+2v)= 2au+ 3av+ bu+ 2bv= (2a+ b)u+ (3a+2b)v showing that it is in the span of {u,v}.

    Conversely, any vector in the span of {u, v} must be of the form au+ bv. To have that of the form x(2u+3v)+ y(u+2v)= (2x+y)u+ (2x+2y)v we must have 2x+ 3y= a and x+ 2y= b. Solving those equations for x and y, we get x= 2a- 3b and y= 2b-a. Actually, we only have to show that we can solve those equations and that can be done by showing that the determinant of the coefficient array, \left|\begin{array}2 & 3 \\ 1 & 2\end{array}\right|= 4-3= 1 and is not 0.
    Last edited by HallsofIvy; March 29th 2009 at 06:26 PM.
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