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Math Help - Find the Eigenvalues and Eigenspaces

  1. #1
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    Find the Eigenvalues and Eigenspaces

    Is my working correct?

    Question: Find the eigenvalues and the corresponding eigenspaces for matrix A:

    A=
    6 -4
    3 -1

    |A - lambda * I| = 0

    6-lambda -4
    3 -1-lambda

    (6-lambda)(-1-lambda) - (3)(-4)

    -6 -6*lambda + lambda + lambda^2 + 12

    6 -5*lambda + lambda^2

    Factorize 6 -5*lambda + lambda^2

    (lambda - 3)(lambda - 2)

    Eigenvalues of A:

    lambda = 3
    lambda = 2


    E(3) =

    6x - 4y = 3x
    3x - y = 3y

    -4y = -3x
    3x = 4y

    Span = ?

    The eigenspaces are ?

    E(2) =

    6x - 4y = 2x
    3x - y = 2y

    -4y = -4x
    3x = 3y

    Span = ?

    The eigenspaces are ?

    Where did I go wrong? How do I find the eigenvalues, what is the "span", and how do I find it?
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  2. #2
    A Plied Mathematician
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    Your work so far is just fine, I would say. You are a bit confused in finding the eigenvectors (they span the eigenspace). Solve the system (A-\lambda I)x=0 for x. Since you have chosen \lambda such that the matrix A-\lambda I is singular, you're guaranteed that the eigenvectors you find are not unique. In fact, any scalar multiple of an eigenvector is an eigenvector. So what do you get when you solve the system (A-\lambda I)x=0 for x?
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  3. #3
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    Thanks for your reply Ackbeet.

    Here is what I got so far...

    (A - lambda I)x = 0

    6-lambda -4
    3 -1-lambda
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  4. #4
    MHF Contributor
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    Quote Originally Posted by sparky View Post


    Eigenvalues of A:

    lambda = 3
    lambda = 2


    E(3) =

    6x - 4y = 3x
    3x - y = 3y

    -4y = -3x
    3x = 4y

    Span = ?

    The eigenspaces are ?

    E(2) =

    6x - 4y = 2x
    3x - y = 2y

    -4y = -4x
    3x = 3y

    Span = ?

    The eigenspaces are ?

    Where did I go wrong? How do I find the eigenvalues, what is the "span", and how do I find it?
    I find it easier to do:

    \displaystyle\lambda=3 \ \ \ \displaystyle\begin{bmatrix}6-3&-4\\3&-1-3\end{bmatrix}\Rightarrow\begin{bmatrix}3&-4\\0&0\end{bmatrix}

    3x_1=4x_2

    3x_2

    \displaystyle x_2\begin{bmatrix}4\\3\end{bmatrix}

    The eigenvector for lamba = 3 is

    \displaystyle \begin{bmatrix}4\\3\end{bmatrix}

    Do the same for lambda = 2. Then your eigenspace will be

    \displaystyle\left\{\begin{bmatrix}4\\3\end{bmatri  x},\begin{bmatrix}\alpha\\ \beta\end{bmatrix}\right\} \ \ \ \alpha,\beta\in\mathbb{R}
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