Please check my Eigenvector solutions.

  1. 1. The problem statement, all variables and given/known data
    Find the characteristic equations, eigenvalues and eigenvector of the following matrix

    2. Relevant equations
    Eigen 2.png


    3. The attempt at a solution
    Eigen 1.png

    Somehow somewhere I think the solution is wrong, based on online Eigenvector calculator on the web. Please do provide me actual answers and solutions. Thanks in advance!
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
    Science Advisor

    Your first eigenvector corresponding to [itex]\lambda= 3[/itex] is correct.

    However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote [itex](A- \lambda I)x= 0[/itex]:
    [tex]\begin{bmatrix}2 & 0 \\ 8 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex]
    which gives
    [tex]\begin{bmatrix}2x_1 \\ 8x_1- 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
    which then requires that [itex]2x_1= 0[/itex] and [itex]8x_1- 2x_2= 0[/itex].
    From the second [itex]x_2= 4x_1[/itex] but the first says [itex]x_1= 0[/itex] so [itex]x_2= 4(0)= 0[/itex]. There is NO non-trivial vector satisfying this. 1 is NOT an eigenvalue
    (I have no idea where you got "[itex]x_1+ x_2= 0[/itex]".)

    The characteristic equation is given by
    [tex]\left|\begin{array}{cc}3- \lambda & 0 \\ 8 & -1- \lambda\end{array}\right|= 0[/tex]
    Which is, of course, simply [itex](3-\lambda)(-1- \lambda)= 0[/itex].
     
  4. Eigen 3.png
    Take a look at this. I've corrected it. Please let this answer correct :)

    Some mistake there;
    Eigenvector for lambda = 1 is [0;1]

    One more, on the second last line.
    is that correct to state "Let x_2=t" or "Let x_1=t"?
     
    Last edited: Aug 12, 2011
  5. If you compute what you think are an eigenvector and eigenvalue pair, stick them back in! They had better satisfy [itex] Ax = \lambda x[/itex] since that is, after all, the equation whose solutions you were looking for in the first place.

    The variable [itex]x_2[/itex] should equal t, if that's what you're asking.
     
  6. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
    Science Advisor

    You mean "for lambda= -1".

    You had just shown that [itex]x_1= 0[/itex] so you can't say "let x_1=t".
     
  7. Thanks Stringy and HallsofIvy :)
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?