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Math Help - Need help, test coming soon!

  1. #1
    Junior Member
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    Exclamation Need help, test coming soon!

    Need explanations!

    My test is coming up real soon!

    Need help with these questions.

    Sorry for the rudeness ... tests make me... nervous...

    Describe all vectors v = [x,y] that are orthogonal to u = [a,b].

    And I really really need help understanding "span" in linear algebra. How do you determine if a vector is the span of a matrix?

    prob) determine if the vector b is in the span of the columns of the matrix A.
    A =
    [1 2]
    [3 4]
    b =
    [5]
    [6]

    prob) show that R^2 =
    span
    ([1] , [1])
    ([1] , [-1])
    Last edited by aeubz; September 23rd 2008 at 07:03 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
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    Chicago, IL
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    Hello

    Since I don't know much about span, I'll take the easy one

    Quote Originally Posted by aeubz View Post
    Describe all vectors v = [x,y] that are orthogonal to u = [a,b].
    Take the dot product of \bold v and \bold u

    \bold u\cdot\bold v=ax+by

    The vectors are orthogonal when \bold u\cdot\bold v=0

    So , we see that ax+by=0\implies by=-ax

    Now, we need to manipulate the vector \bold v:

    \left[\begin{array}{c}x\\y\end{array}\right]={\color{red}\frac{b}{b}}\left[\begin{array}{c}x\\y\end{array}\right]=\frac{1}{b}\left[\begin{array}{c}bx\\by\end{array}\right]

    Since, by=-ax, \bold v=\frac{1}{b}\left[\begin{array}{c}bx\\-ax\end{array}\right]=\frac{x}{b}\left[\begin{array}{c}b\\-a\end{array}\right]

    Since \frac{x}{b} is another constant, let's denote it by k

    So all the possible vectors will take on the form \color{red}\boxed{\bold v=k\left[\begin{array}{c}b\\-a\end{array}\right];~k,a,b\in\mathbb{R}}.

    Does this make sense?

    --Chris
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