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Math Help - Basis perpendicular to a subspace spanned by 2 vectors

  1. #1
    Member mybrohshi5's Avatar
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    Basis perpendicular to a subspace spanned by 2 vectors

    Find a basis perpendicular to W.

    Let  V = \begin{bmatrix}1\\-4\\1\\4\end{bmatrix} and  U =  \begin{bmatrix}3\\-13\\3\\-4\end{bmatrix}

    and let W be the subspace spanned by V and U.

    I am pretty confused on this topic right now I just dont seem to know what i need to do to solve problems like these.

    Any help would be great

    Thank you
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  2. #2
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    Quote Originally Posted by mybrohshi5 View Post
    Find a basis perpendicular to W.

    Let  V = \begin{bmatrix}1\\-4\\1\\4\end{bmatrix} and  U =  \begin{bmatrix}3\\-13\\3\\-4\end{bmatrix}

    and let W be the subspace spanned by V and U.

    I am pretty confused on this topic right now I just dont seem to know what i need to do to solve problems like these.

    Any help would be great

    Thank you
    Do you want an orthogonal basis or orthonormal basis?
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  3. #3
    Member mybrohshi5's Avatar
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    The question doesnt specify. It just says....... let W the subspace of  {\mathbb R}^4 spanned by v and u. Find a basis of  W^{\perp} .

    I should probably know how to find both though im sure
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  4. #4
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    If you can tell me what the answer is, I can tell you what one it is looking for.
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  5. #5
    Member mybrohshi5's Avatar
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    Im not sure what the answer is. Its for online homework.

    There are two blank 4x1 matrixes where i can put my answer in. so it looks like this

     \begin{bmatrix}.\\.\\.\\.\end{bmatrix}, \begin{bmatrix}.\\.\\.\\.\end{bmatrix}
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  6. #6
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    Both would be of that form.

    Use the Gram-Schmidt process.

    Do you know it?
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  7. #7
    Member mybrohshi5's Avatar
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    Quote Originally Posted by dwsmith View Post
    Both would be of that form.

    Use the Gram-Schmidt process.

    Do you know it?
    Yes i do. Is that all i have to do? Seems easy enough.

    and will the Gram schmidt process give me an orthogonal basis correct? then from there i could find a orthonormal basis if i needed right?
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  8. #8
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    You can normalize it if you would like.
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  9. #9
    Member mybrohshi5's Avatar
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    Using the gram schmidt process is not getting me the right answer

    I even normalized the orthogonal basis i got and those are wrong as well.

    Using the process i got

    \begin{bmatrix}1\\-4\\1\\4\end{bmatrix} and \begin{bmatrix}\frac{30}{17}\\-\frac{137}{17}\\\frac{30}{17}\\-\frac{152}{17}\end{bmatrix}

    i plugged those in and they were wrong.

    so i normalized those to get the orthononormal vectors

    \frac{1}{\sqrt(34)}*\begin{bmatrix}1\\-4\\1\\4\end{bmatrix} and \frac{1}{\sqrt(2569/17)}*\begin{bmatrix}\frac{30}{17}\\-\frac{137}{17}\\\frac{30}{17}\\-\frac{152}{17}\end{bmatrix}

    Can you see where i am going wrong?
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  10. #10
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    I just checked my dot product and it wasn't zero but your dot product is. What section in your book does this question relate too?
    Last edited by dwsmith; April 18th 2010 at 10:28 AM.
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  11. #11
    Member mybrohshi5's Avatar
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    This is for online homework and i dont have the book cause i am fairly good at math and the professor said we didnt need it unless we needed an extra source of help (mine is this forum haha).

    If i look at the professors power points i would think it relates to the section called Orthogonal Vectors in R^n, but i am not positive....
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  12. #12
    Member mybrohshi5's Avatar
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    I found this on yahoo answers and it looks similar to my question but i am a little unsure what is done to get the answer. maybe you will know what is going on

    Find a basis of the subspace of R4 that consists of all vectors perpendicular to both:? - Yahoo! Answers
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  13. #13
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    I got these
    (-68, -16, 0, 1)
    ( -1, 0, 1, 0).
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  14. #14
    Member mybrohshi5's Avatar
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    Quote Originally Posted by zzzoak View Post
    I got these
    (-68, -16, 0, 1)
    ( -1, 0, 1, 0).
    What did you do to find those?
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  15. #15
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    When I followed that example, I obtained x_3\begin{bmatrix}<br />
-1\\ <br />
0\\ <br />
1\\ <br />
0<br />
\end{bmatrix}+x_4\begin{bmatrix}<br />
-68\\ <br />
-16\\ <br />
0\\ <br />
1<br />
\end{bmatrix}
    Last edited by dwsmith; April 19th 2010 at 05:19 PM.
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