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Math Help - eigenvectors

  1. #1
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    eigenvectors

    Find a basis of the linear space V of all 2x2 matrices A for which
    [1; -3] is an eigenvector, and thus determine the dimension of V (; denotes a new row).

    Attempt at a solution:
    So A[1; -3] = (lambda)[1; -3].
    [a b; c d][1; -3] = (lambda)[1; -3]. Here I'm stuck--is this even the right direction to go in?
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  2. #2
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    Quote Originally Posted by noles2188 View Post
    Find a basis of the linear space V of all 2x2 matrices A for which
    [1; -3] is an eigenvector, and thus determine the dimension of V (; denotes a new row).

    Attempt at a solution:
    So A[1; -3] = (lambda)[1; -3].
    [a b; c d][1; -3] = (lambda)[1; -3]. Here I'm stuck--is this even the right direction to go in?
    Yes, that's a good direction. Now do the multiplication on the left to get two equations is a, b, c, d, and \lambda.

    You should get a- 3b= \lambda and c- 3d= -3\lambda. We need to get rid of \lambda to have a condition on a, b, c, and d only so I would try dividing one equation by the other: \frac{c-3d}{a-3b}= -3 so c-3d= -3a+ 9b. You can choose a value of any 3 of those and solve for the remaining one. You can find a basis by taking each of the 3 values equal to 1 in turn and the other two 0.
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  3. #3
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    Quote Originally Posted by noles2188 View Post
    Find a basis of the linear space V of all 2x2 matrices A for which
    [1; -3] is an eigenvector, and thus determine the dimension of V (; denotes a new row).

    Attempt at a solution:
    So A[1; -3] = (lambda)[1; -3].
    [a b; c d][1; -3] = (lambda)[1; -3]. Here I'm stuck--is this even the right direction to go in?

    Yes, the direction looks right....let us continue a little more:

    \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\  left(\begin{array}{c}1\\\!\!\!\!-3\end{array}\right)=\lambda\left(\begin{array}{c}1  \\\!\!\!\!-3\end{array}\right)\Longrightarrow \left(\begin{array}{c}a-3b\\c-3d\end{array}\right)=\left(\begin{array}{c}\lambda  \\\!\!\!-3\lambda\end{array}\right) \Longrightarrow\,3\lambda=3a-9b=3d-c=3\lambda\Longrightarrow\,3a-9b-3d+c=0

    So any matrix \left(\begin{array}{cc}x&y\\w&z\end{array}\right) whose (ordered!) entries belong to the hyperplane \left\{\left(\begin{array}{c}x\\y\\z\\w\end{array}  \right)\in \mathbb{R}^4\,\,\slash \,3x-9y-3z+w=0\right\} will have the given vector \left(\begin{array}{c}1\\\!\!\!\!-3\end{array}\right) as eigenvector
    wrt the eigenvalue \lambda=a-3b

    Tonio
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