Results 1 to 2 of 2

Thread: Eigenvectors/values

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    56

    Eigenvectors/values

    Let A = $\displaystyle \begin{bmatrix}1&-1&0\\-1&1&0\\0&0&1\end{bmatrix}$ and $\displaystyle T: R^3 \rightarrow R^3$ by T(X) = AX.
    Find eigenpairs for A, find a basis for Ker(T) using the eigenvectors previously found, find a basis for Im(T) using eigenvectors previously found.

    The eigenpairs I got were...
    (0, $\displaystyle \begin{bmatrix}1\\1\\0\end{bmatrix}$), (1, $\displaystyle \begin{bmatrix}0\\0\\1\end{bmatrix}$), and (2, $\displaystyle \begin{bmatrix}-1\\1\\0\end{bmatrix}$

    I wasn't sure about the Kernel or Image parts but for the kernel I got
    t$\displaystyle \begin{bmatrix}1\\1\\0\end{bmatrix}$
    And for the image I got
    $\displaystyle [r\begin{bmatrix}0\\0\\1\end{bmatrix}, s\begin{bmatrix}-1\\1\\0\end{bmatrix}]$

    Could someone please verify if this is correct or not? I assume the kernel and image are not so could someone also please help me with that?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,780
    Thanks
    3028
    Yes, that is correct. Your matrix is 3 by 3 so the base space is 3 dimensional and you have three independent eigenvectors. Since there are 3 and they are independent (eigevectors corresponding to distinct eigenvalues are always independent), they form a basis for the space. That means that any vector v, can be written as $\displaystyle a\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$.

    Applying A to that, $\displaystyle Av= A(a\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix})$$\displaystyle = aA\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ bA\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ cA\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$$\displaystyle = a(0)\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b(1)\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c(2)\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$.

    That is, Av= 0 if and only if $\displaystyle v= a\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ so $\displaystyle \{\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}\}$ is a basis for the kernel of A and $\displaystyle Av= b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ 2c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$ so $\displaystyle \{\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, \begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}\}$ is a basis for the image.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Predicting maximum values from minimum values
    Posted in the Statistics Forum
    Replies: 0
    Last Post: Oct 4th 2011, 07:42 AM
  2. Eigenvectors/values Question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jun 15th 2011, 01:13 PM
  3. Replies: 1
    Last Post: Jun 1st 2011, 01:47 AM
  4. Eigenvectors and Values for T(A)=A^t
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: May 13th 2010, 03:43 PM
  5. Replies: 1
    Last Post: May 24th 2009, 05:16 AM

Search Tags


/mathhelpforum @mathhelpforum