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Thread: 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 $\displaystyle R^3$

    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:

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

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

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

    $\displaystyle \lambda_1=0$:

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

    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 $\displaystyle \lambda_2,\lambda_3=1$:

    z=0,

    Which gives: $\displaystyle {(x,y,0)}$, and implies: $\displaystyle {(1,0,0),(0,1,0)}$, 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; Dec 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:
    $\displaystyle \lambda_2,\lambda_3=1$
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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 $\displaystyle T:\mathbb{R}^2\longrightarrow{\mathbb{R}^2}$ its known that $\displaystyle \vec{v}=(1,1)$ its an eigenvector with the eigenvalue associated $\displaystyle \lambda=2$ and that $\displaystyle T(0,1)=(1,2)$.

    Find $\displaystyle [T]_c$

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

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

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

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

    $\displaystyle
    \begin{bmatrix}{1}&{1}\end{bmatrix}\begin{bmatrix} {x}\\{y}\end{bmatrix} = \begin{bmatrix}x + y\end{bmatrix}
    $
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