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Math Help - Eigenvalues and eigenvectors [Linear Algebra]

  1. #1
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    Eigenvalues and eigenvectors [Linear Algebra]

    Hi there. I must give the eigenvalues and the eigenvectors for the matrix transformation of the orthogonal projection over the plane XY on R^3

    So, at first I thought it should be the eigenvalue 1, and the eigenvectors (1,0,0) and (0,1,0), because they don't change. But I also tried doing the calculus, and then I've confused.

    I have:

    [T]_c=\begin{bmatrix}{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0  }&{0}\end{bmatrix}

    From the characteristic polynomial I get to: -\lambda(1-\lambda)^2

    Then, I have as eigenvalues 0, and 1 twice.

    \lambda_1=0:

    I get to: \begin{Bmatrix}x=0\\y=0\end{matrix}
    and then the eigenvector: {(0,0,1)}

    So I thought, shouldn't it be zero? because of the projection. I have doubts with this, but I know that as it is a symmetrical matrix it should be diagonalizable, and then I should get a basis from the eigenvectors, which I wouldn't find with just the first reasoning, and then I need a linear independent vector, like this one, respect to the first I gave.

    And then for \lambda_2,\lambda_3=1:

    z=0,

    Which gives: {(x,y,0)}, and implies: {(1,0,0),(0,1,0)}, I think that have sense.

    Well, I need some help with this. Can anybody tell me if this is right, and if it isn't, what I did wrong?

    Thank you!

    Bye there.

    PS: I have my final exam tomorrow :P
    Last edited by Ulysses; December 28th 2010 at 09:22 AM.
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  2. #2
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    Yes, you are correct.
    3 eigen values 0, 1, 1.

    Projection collapses every vector parallel to the z axis to the zero vector. (0,0,a) is transformed to 0*(0,0,a) = (0,0,0), so its an eigen vector right?

    And I think you mean:
    \lambda_2,\lambda_3=1
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  3. #3
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    You're right. Thanks.
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  4. #4
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    I have another doubt about this. Now its about the construction of a transformation matrix. The thing is I thought I was constructed in one way, and it seems to be that I was wrong.

    I have this problem, which says: From a linear transformation T:\mathbb{R}^2\longrightarrow{\mathbb{R}^2} its known that \vec{v}=(1,1) its an eigenvector with the eigenvalue associated \lambda=2 and that T(0,1)=(1,2).

    Find [T]_c

    I thought of [T]_c as T(x,y)=(x+y,2y)

    The thing is that when I constructed the matrix I did it this way:

    [T]_c=\begin{bmatrix}{1}&{0}\\{1}&{2}\end{bmatrix}

    But when I make the multiplication with the vectors that the problem data gives, I don't get the right results, but when I changed to [T]_c=\begin{bmatrix}{1}&{1}\\{0}&{2}\end{bmatrix} it seems to be the matrix I'm looking for.

    So, it seems to be that x, and y go in the columns, and not in the rows as I thought. I was wrong then?
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  5. #5
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    Yes, you are correct. x and y coefficients go in the column.

    If you forget think about it this way.

    <br />
\begin{bmatrix}{1}&{1}\end{bmatrix}\begin{bmatrix}  {x}\\{y}\end{bmatrix} = \begin{bmatrix}x + y\end{bmatrix}<br />
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