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Thread: Linear Algebra understanding question

  1. #1
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    Linear Algebra understanding question

    find the vector $\displaystyle \vec{v}\in \mathbb{R}^3$
    who meets the requirements:

    for every matrix $\displaystyle A=\begin{bmatrix}
    a_1_1 & a_1_2 & a_1_3\\
    a_2_1 & a_2_2 & a_2_3\\
    a_3_1 & a_3_2 & a_3_3
    \end{bmatrix}$

    exist:
    $\displaystyle A\vec{v}=\begin{bmatrix}
    a_1_2-a_1_3 \\
    a_2_2-a_2_3 \\
    a_3_2-a_3_3
    \end{bmatrix}$

    Can some one please help me here? is $\displaystyle \vec{v}$ is eigenvector?!
    Thank you!
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  2. #2
    Senior Member
    Joined
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    Define your vector to be
    $\displaystyle
    \vec{v}=\begin{bmatrix}
    x \\
    y \\
    z
    \end{bmatrix}
    $

    And look at the value of $\displaystyle A\vec{v}$ algebraically. You should be able to get exactly what $\displaystyle \vec{v}$ should be.

    For example, the first row of $\displaystyle A\vec{v}$ should be $\displaystyle a_{11}x + a_{12}y + a_{13}z$, and this is equal to $\displaystyle a_{12} - a_{13}$.
    What are $\displaystyle x, y, z$?
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  3. #3
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    it seems to me that is should be $\displaystyle \begin{bmatrix}
    0\\
    1\\
    -1
    \end{bmatrix}$
    am i right?

    and i cold just write it like that?
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  4. #4
    Senior Member
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    Dec 2010
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    Yes, that is the only possible vector for the solution.

    If you are unsure pick values for A that make A invertible (easiest would be the identity matrix).

    Now substituting values, we have:

    $\displaystyle
    \begin{bmatrix}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 1
    \end{bmatrix}
    \vec{v}=\begin{bmatrix}
    0 - 0 \\
    1 - 0 \\
    0 - 1
    \end{bmatrix}
    $
    Which gives you the unique solution.
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  5. #5
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    Ok i got it!
    Thank you very much!
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