Results 1 to 3 of 3

Math Help - Eigenspaces, Orthogonal + Diagonal Matrices

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    1

    Eigenspaces, Orthogonal + Diagonal Matrices

    Could someone please let me know how I would go about accomplishing the following tasks. Thank you.



    Let A = [[4,0,0],[0,1,3],[0,3,1]

    a) Find the Eigenvalues of A.

    (lambda - 4)(lambda - 4)(lambda + 2)
    So the eigenvalues are 4, 4, and -2.

    b) For each eigenvalue, find a basis for the corresponding eigenspace.

    ??

    c) Use these bases vectors to construct an orthogonal matrix P.

    ??

    d) Verify that P^-1AP is a diagonal matrix.

    ??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by kabirSarin View Post
    Could someone please let me know how I would go about accomplishing the following tasks. Thank you.



    Let A = [[4,0,0],[0,1,3],[0,3,1]

    a) Find the Eigenvalues of A.

    (lambda - 4)(lambda - 4)(lambda + 2)
    So the eigenvalues are 4, 4, and -2.

    b) For each eigenvalue, find a basis for the corresponding eigenspace.

    ??

    c) Use these bases vectors to construct an orthogonal matrix P.

    ??

    d) Verify that P^-1AP is a diagonal matrix.

    ??
    Eigenvalues of A are

    \begin{bmatrix}<br />
4-\lambda & 0 & 0\\ <br />
0 & 1-\lambda & 3\\ <br />
0 & 3 & 1-\lambda<br />
\end{bmatrix}\rightarrow (4-\lambda)[(1-\lambda)^2-9]=-(\lambda-4)^2(\lambda+2)

    When \lambda=4

    \begin{bmatrix}<br />
0 & 0 & 0\\ <br />
0 & -3 & 3\\ <br />
0 & 3 & -3<br />
\end{bmatrix}\rightarrow rref= \begin{bmatrix}<br />
0 & 0 & 0\\ <br />
0 & 1 & -1\\ <br />
0 & 0 & 0<br />
\end{bmatrix}

    x_1
    x_2=x_3
    x_3

    \begin{bmatrix}<br />
x_1\\ <br />
x_3\\ <br />
x_3<br />
\end{bmatrix}\rightarrow x_1\begin{bmatrix}<br />
1\\ <br />
0\\ <br />
0<br />
\end{bmatrix}+x_3\begin{bmatrix}<br />
0\\ <br />
1\\ <br />
1<br />
\end{bmatrix}

    Now find the other eigenvector for lambda = -2

    Once you have the third eigenvector, you can construct your P matrix that corresponds to A
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,429
    Thanks
    1858
    I prefer to use the definition of "eigenvector" directly: If 4 is an eigenvector for A, then there exist a vector, v, such that Av= 4v. Specifically,
    \begin{bmatrix}4 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 3 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}4x \\ y+ 3z\\ 3y+ z\end{bmatrix}= 4\begin{bmatrix} x \\ y \\ z\end{bmatrix}.

    That gives the four equations 4x= 4x, y+ 3z= 4y, and 3y+ z= 4z. The first equation is satisfied by any x, of course and the last two are the same as 3y= 3z or z= y. That is, any eigenvector, corresponding to eigenvalue 4, is of the form \begin{bmatrix}x \\ y \\ y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}, showing the two basis vectors for the eigenspace clearly.

    With eigenvalue -2, those equations are simply 4x= -2x, y+ 3z= -2y, and 3y+ z= -2z. The first equation says x= 0 and I will leave the others to you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 15th 2011, 05:32 AM
  2. General solution for diagonal zero matrices?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 3rd 2010, 04:54 PM
  3. Field of diagonal matrices (or not...)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: June 30th 2010, 12:48 PM
  4. Linear maps and diagonal matrices
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 5th 2010, 12:23 PM
  5. Diagonal Matrices from Characteristic Poly
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 7th 2008, 12:57 PM

Search Tags


/mathhelpforum @mathhelpforum