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A question about finding vectors

  1. Jan 29, 2008 #1
    i added a link with the question and how i tried to solve it

    i want to create a matrix using the vectors that i found

    the problem is
    that i got 4 vectors and two of them is (0,0,0)

  2. jcsd
  3. Jan 29, 2008 #2


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    This is a problem about eigenvalues and eigenvectors, right?

    Well, (0, 0, 0) is never an eigenvector since the whole point of an "eigenvalue", [itex]\lambda[/itex], of T is that there exist a non-zero vector v such that [itex]Tv= \lambda v[/itex].

    You don't clearly state what the original matrix is but I think you mean
    [tex]M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right][/tex]

    The characteristic equation for that is not at all what you give, however. I get [itex]-\lambda^3+ 3\lambda^2= 0[/itex] which has roots 0, 0, 3. That makes sense to me since the matrix is badly "singular". It's no surprise that it has 0 as a double eigenvalue. To find eigenvectors corresponding to the eigenvalue 0, we look at
    [tex]M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right][/tex]

    Obviously that gives us the single equation x+ y+ z= 0. The "solution space" is two dimensional- we can pick any values for x and y and solve for z. In particular if we take x= 1, y= 0, z= -1. An eigenvector corresponding to eigenvalue 0 is [1 0 -1]T. If we choose x= 0, y= 1, we get z= -1. An independent eigenvector corresponding to eigenvalue 0 is [0 1 -1]T.

    To find an eigenvector corresponding to eigenvalue 3, we look at the equation
    [tex]M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 3x \\ 3y \\ 3z \end{array}\right][/tex]
    That gives three equations, x+ y+ z= 3x, x+ y+ z= 3y, and x+ y+ z= 3z which are, again, dependent equations. This time they reduce to x= y= z. An eigenvector corresponding to eigenvalue 3 is [1 1 1]T.

    The matrix you are looking for, that diagonalizes M, is
    [tex]M=\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 &1 \\ -1 & -1 & 1\end{array}\right][/tex]
    Last edited: Jan 29, 2008
  4. Jan 29, 2008 #3

    how to prove that this matrix is orthogonal
    and if not
    how to transform it to an orthogonal matrix

    Last edited: Jan 29, 2008
  5. Jan 30, 2008 #4


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    See if it satisfies your definition of orthogonal, of course! There are several equivalent definitions (ATA= I is one) and I don't know which you were given. However, either as a definition, or as a theorem, it is true that the individual columns of an orthogonal matrix, as vectors, form an orthonormal basis for the vector space: the dot product of two different columns must be 0 and the dot product of a column with itself must be 1. Here, I made no attempt to reduce those vectors to an orthonormal basis:
    [1 0 -1], [0 1 -1], and [1 1 1] happen to be "orthogonal" (their dot products are 0) but are not normalized. (One can show that eigenvectors corresponding to different eigenvalues must be orthogonal. The two eigenvectors corresponding to 0 were not necessarily orthogonal, but my choice of x=1 y=0 and x=0 y=1 forced that.) Find the length of each vector and divide each vector by its length. Using those "normalized" eigenvectors as columns will give you an orthogonal matrix.
  6. Feb 3, 2008 #5
  7. Feb 4, 2008 #6
    i have solved it again
    but i dont get them to be orthogonal
    what to do in that case??

    i didnt get the dot product zero
    (so its not orthogonal)

    what to do??

    and i have writen bellow what what values i need to divide each vector in order to make
    the matrix orthonormal
    did i make it ok??

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