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Math Help - Find subspace

  1. #1
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    Find subspace

    Find the basis for the solution space in R^5 of this linear system. Find this subspace dimension:

    [1 2 3 4 5
    2 -1 2 -1 2
    0 5 4 9 8
    1 7 7 13 13]

    ;; 1x+ 2y + 3z + 4s +5d = 0
    ;; 2x + -1y + 2z + -1s + 2d = 0
    ;; 0x + 5y + 4z + 9s + 8d = 0
    ;; 1x + 7y + 7z + 13s + 13d = 0

    row reduction =

    1 0 7/5 2/5 9/5

    0 1 4/5 9/5 8/5

    0 0 0 0 0

    0 0 0 0 0



    so x3 x4 x5 are free

    x1 = -7/5 x3 - 2/5 x4 - 9/5 x5
    x2 = -4/5 x3 - 9/5 x4 - 8/5 x5

    therefore linearly independent and form a basis for their span of 3 dimensions
    is this correct ?
    Last edited by harveyo; January 11th 2010 at 09:56 PM.
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  2. #2
    Member Mollier's Avatar
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    Elimination gives me:

    <br />
\begin{array}[pos]{ccccc}<br />
    1 & 2 & 3 & 4 & 6\\<br />
    0 & -5&-4 &-9 &-10\\<br />
    0 & 0 & 0 & 0 &-2\\<br />
    0 & 0 & 0 & 0 & 0<br />
\end{array}<br />

    could you check your work (and perhaps mine) ?

    You also write :
    ;; 1x+ 2y + 3z + 4s +5d = 0

    while it is 1 2 3 4 6 in your "matrix"..
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  3. #3
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    Quote Originally Posted by Mollier View Post
    Elimination gives me:

    <br />
\begin{array}[pos]{ccccc}<br />
1 & 2 & 3 & 4 & 6\\<br />
0 & -5&-4 &-9 &-10\\<br />
0 & 0 & 0 & 0 &-2\\<br />
0 & 0 & 0 & 0 & 0<br />
\end{array}<br />

    could you check your work (and perhaps mine) ?

    You also write :
    ;; 1x+ 2y + 3z + 4s +5d = 0

    while it is 1 2 3 4 6 in your "matrix"..
    sorry it should be a 5
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Mollier View Post
    Elimination gives me:

    <br />
\begin{array}[pos]{ccccc}<br />
1 & 2 & 3 & 4 & 6\\<br />
0 & -5&-4 &-9 &-10\\<br />
0 & 0 & 0 & 0 &-2\\<br />
0 & 0 & 0 & 0 & 0<br />
\end{array}<br />

    could you check your work (and perhaps mine) ?

    You also write :
    ;; 1x+ 2y + 3z + 4s +5d = 0

    while it is 1 2 3 4 6 in your "matrix"..
    When the 5 - 6 confussion is corrected and the arithmetic fixed accordingly this is essentially correct. There are two free variables and so the solution space is two dimensional.

    CB
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  5. #5
    Member Mollier's Avatar
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    harveyo: Perhaps you could re-do the elimination now?
    You could also try looking at how I posted my matrix, and copy-paste it into your own post. It makes it easier to read.
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  6. #6
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    i feel my elimination is correct! i do not know how to post the array properly
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  7. #7
    Member Mollier's Avatar
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    You can click on the matrix I wrote and the code will pop up in a new window.

    It looks like this:

    [--math]
    \begin{array}[pos]{ccccc}
    1 & 2 & 3 & 4 & 6\\
    0 & -5&-4 &-9 &-10\\
    0 & 0 & 0 & 0 &-2\\
    0 & 0 & 0 & 0 & 0
    \end{array}
    [/math--]

    Without the "--" in  and .

    Now you just copy-paste that into your own reply and it should work. Give it a try.
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