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Thread: 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 $\displaystyle (A-\lambda I)x=0$ for $\displaystyle x.$ Since you have chosen $\displaystyle \lambda$ such that the matrix $\displaystyle A-\lambda I$ 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 $\displaystyle (A-\lambda I)x=0$ for $\displaystyle x?$
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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:

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

    $\displaystyle 3x_1=4x_2$

    $\displaystyle 3x_2$

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

    The eigenvector for lamba = 3 is

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

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

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