Please check my Eigenvector solutions.

  1. Aug 12, 2011 #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. Aug 12, 2011 #2

    HallsofIvy

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    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. Aug 12, 2011 #3
    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. Aug 12, 2011 #4
    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. Aug 12, 2011 #5

    HallsofIvy

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    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. Aug 13, 2011 #6
    Thanks Stringy and HallsofIvy :)
     
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