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Rowspace and kernel

  1. Apr 18, 2006 #1
    [tex]A = \left(\begin{array}{cccc}-1 &6&5&9 \\ -1&0&1&3 \end{array}\right)[/tex]

    Find orthonormal bases of the kernel, row space.

    To find the bases, I did reduced the array to its RREF.

    [tex]A = \left(\begin{array}{cccc}1 & 0&-1&-3\\ 0&1&2/3&1 \end{array}\right)[/tex]

    Then the orthonormal bases would just be that divided by the length.

    [tex]||v_1||=\sqrt{1+1+3^2}=\sqrt{11}[/tex]

    [tex]||v_2||=\sqrt{1+(2/3)^2+1}=\sqrt{2.44444}[/tex]

    so that means, the orthonormal bases would be:

    [tex]A = \left(\begin{array}{cccc} \frac{1}{ \sqrt{11}} & 0&\frac{-1}{ \sqrt{11}}&\frac{-3}{ \sqrt{11}} \\0 & \frac{1}{ \sqrt{2.44444}} & \frac{.66666}{ \sqrt{2.44444}} &\frac{1}{ \sqrt{2.44444}}\end{array}\right)[/tex]

    what exactly is the orthonormal bases of the kernel?
    Also, isnt the row space the same as the vectors of the bases?
    I think I also did something wrong in my calculations
     
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 18, 2006 #2

    0rthodontist

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    If you are looking for an orthonormal basis, one thing to check is that your basis actually is orthogonal. The basis you have found for the row space is not orthogonal.

    Given an arbitrary basis for a vector space, do you know how to construct an orthogonal basis for it via the Gram-Schmidt process?
     
  4. Apr 19, 2006 #3
    is what i did above the orthonormal row space? that is wrong as well, i dont know why.... however:

    using the Gram-Schmidt process, i still get an error:

    [tex]A = \left(\begin{array}{cccc}-1 &6&5&9 \\ -1&0&1&3 \end{array}\right)=\left(\begin{array}{cc}W_1 &W_2 \end{array}\right)[/tex]

    want to find an orthonormal basis [tex]R={U_1 ,U_2}[/tex]

    [tex]U_1=\frac{W}{||W_1||}[/tex]
    [tex]||W_1||=\sqrt{11}[/tex]
    [tex]U_1=\frac {\left(\begin{array}{cccc}-1 &6&5&9 \end{array}\right)}{\sqrt{11}}[/tex]

    [tex]U_2=\frac{W_2-<W_2 , U_1 > U_1}{||W_2-<W_2 , U_1 > U_1||}[/tex]

    where [tex]W_2=\left(\begin{array}{cccc} -1&0&1&3 \end{array}\right)[/tex]

    [tex]U_1=\frac {\left(\begin{array}{cccc}-1 &6&5&9 \end{array}\right)}{\sqrt{11}}[/tex]
    [tex]U_1=\left(\begin{array}{cccc}-1/\sqrt{11} &6\sqrt{11}&5\sqrt{11}&9\sqrt{11} \end{array}\right)[/tex]

    is the the correct set up to get an orthonormal basis?

    Also, to get an orthonormal row space, what would I have to do?
     
    Last edited: Apr 19, 2006
  5. Apr 19, 2006 #4

    0rthodontist

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    A is not equal to (W1 W2). W1 and W2 are a basis for the row space of A, they are not equal to A when written like that because they are horizontal.

    It looks like the method you have set up to find U1 and U2 is correct. What do you mean an "orthonormal row space"? You can find an orthonormal _basis_ U1, U2 for the row space just by finding the vector U2 according to the formula you've set up.
     
  6. Apr 20, 2006 #5
    I typed my W1 and W2 vectors wrong. W1 is supposed to represent the vector (-1 6 5 9) and W2 = (-1 0 1 3)

    What do I get when I find U1 and U2? Is U1 and U2 the row space? or is U1 an U2 the orthonormal basis?
     
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