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Math Help - Question about eigenvectors(eigenspaces)

  1. #1
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    Question about eigenvectors(eigenspaces)

    The question asks to find the solution of the system x'(t)=Ax(t)

    given my A is
    \begin{bmatrix}<br />
  -7 & 0 & 6 \\<br />
  0 & 5 & 0 \\<br />
6 & 0 & 2 <br />
\end{bmatrix}<br />

    After finding my det( \lambda I -A)=> \lambda ^3 -75\lambda +250 =0

    I found three real roots:
    \lambda 1=-10
    \lambda 2=5
    \lambda 3=5

    Now starting with the first root equal to -10 and plugging into the det matrix
    \begin{bmatrix}<br />
  -3 & 0 & -6 \\<br />
  0 & -15 & 0 \\<br />
-6 & 0 & -12 <br />
\end{bmatrix}<br />
    *
    \begin{bmatrix}<br />
  x\\<br />
  y\\<br />
z <br />
\end{bmatrix}<br />
    =
    \begin{bmatrix}<br />
  0\\<br />
  0\\<br />
0 <br />
\end{bmatrix}<br />

    I found my U1 to be <2, 0, -1>

    Now I'm confused what to do with the last two lambdas, I get the feeling that there should be more than just two Cn's
    x(t)=C1 e^-10t\begin{bmatrix}<br />
  2\\<br />
  0\\<br />
1 <br />
\end{bmatrix}+C2...+C3...<br />

    My question is can someone show me how to solve for u2 and u3 (if there is a u3), the way my textbook explains solving for eigenvectors is not helping very much considering the questions are getting harder and the method I've been using doesn't seem to be working.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by 360modina View Post
    The question asks to find the solution of the system x'(t)=Ax(t)

    given my A is
    \begin{bmatrix}<br />
  -7 & 0 & 6 \\<br />
  0 & 5 & 0 \\<br />
6 & 0 & 2 <br />
\end{bmatrix}<br />

    After finding my det( \lambda I -A)=> \lambda ^3 -75\lambda +250 =0

    I found three real roots:
    \lambda 1=-10
    \lambda 2=5
    \lambda 3=5

    Now starting with the first root equal to -10 and plugging into the det matrix
    \begin{bmatrix}<br />
  -3 & 0 & -6 \\<br />
  0 & -15 & 0 \\<br />
-6 & 0 & -12 <br />
\end{bmatrix}<br />
    *
    \begin{bmatrix}<br />
  x\\<br />
  y\\<br />
z <br />
\end{bmatrix}<br />
    =
    \begin{bmatrix}<br />
  0\\<br />
  0\\<br />
0 <br />
\end{bmatrix}<br />

    I found my U1 to be <2, 0, -1>

    Now I'm confused what to do with the last two lambdas, I get the feeling that there should be more than just two Cn's
    x(t)=C1 e^-10t\begin{bmatrix}<br />
  2\\<br />
  0\\<br />
1 <br />
\end{bmatrix}+C2...+C3...<br />

    My question is can someone show me how to solve for u2 and u3 (if there is a u3), the way my textbook explains solving for eigenvectors is not helping very much considering the questions are getting harder and the method I've been using doesn't seem to be working.
    When you have a matrix equation of the form \frac{\,d\bold x}{\,dt}=\bold A\bold x, where \bold A is the coefficient matrix, we assume the solution to be of the form \bold x=\bold ve^{\lambda t}

    After substitutions and simplifications, we end up with \left(\bold A-\lambda\bold I\right)\bold v=\bold 0, where \bold v is the eigenvector associated with the eigenvalue \lambda

    To find the eigenvalues, we must find where \det\left(\bold A-\lambda\bold I\right)=0

    Now to your question

    Your coefficient matrix is \bold A=\begin{bmatrix}-7 & 0 & 6 \\ 0 & 5 & 0 \\ 6 & 0 & 2\end{bmatrix}

    Thus, to find the eigenvalues, they must satisfy the equation \det\left(\bold A-\lambda\bold I\right)=0

    Thus, we see that

    \left|\begin{array}{ccc}-7-\lambda & 0 & 6 \\ 0 & 5-\lambda & 0 \\ 6 & 0 & 2-\lambda\end{array}\right|=0\implies (-7-\lambda)\left|\begin{array}{cc}5-\lambda & 0 \\0 & 2-\lambda\end{array}\right|+6\left|\begin{array}{cc}  0 & 5-\lambda\\ 6 & 0\end{array}\right|=0

    \implies (-7-\lambda)(5-\lambda)(2-\lambda)-36(5-\lambda)=0\implies (5-\lambda)\left[(-7-\lambda)(2-\lambda)-36\right]=0

    \implies (5-\lambda)\left[-50+5\lambda+\lambda^2\right]=0\implies(5-\lambda)(10+\lambda)(-5+\lambda)=0

    \implies\begin{array}{rcr}\lambda_1&=&5\\\lambda_2  &=&5\\\lambda_3&=&-10\end{array}

    We the case of repeated eigenvalues.

    The eigenvector for \lambda = -10 would be found by solving the equation \begin{bmatrix}3 & 0 & 6 \\ 0 & 15 & 0 \\ 6 & -5 & 12\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bm  atrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

    You end up with the equations 3v_1+6v_3=0, 15v_2=0 and 6v_1-5v_2+12v_3=0

    Since v_2=0, we see that -\tfrac{1}{2}v_1=v_3

    Thus, we can let the eigenvector be \bold v_3=\begin{bmatrix}2\\0\\-1\end{bmatrix}

    Now, let's look at what happens when \lambda = 5

    We have the equation

    \begin{bmatrix}-12 & 0 & 6 \\ 0 & 0 & 0 \\ 6 & -5 & -3\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bma  trix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

    We come up with the equations -12v_1+6v_3=0\implies 2v_1-v_3=0 and 6v_1-5v_2-3v_3=0

    Letting v_2=0, we end up with 6v_1-3v_3=0\implies 2v_1-v_3=0\implies v_3=2v_1

    We can have one of the eigenvectors being \bold v_1=\begin{bmatrix}2\\0\\4\end{bmatrix}

    However, I'm not able to see another possible eigenvector!!!

    Up to this point, our solution is \bold x(t)=c_1\begin{bmatrix}2\\0\\4\end{bmatrix}e^{5t}+  c_2\begin{bmatrix}2\\0\\-1\end{bmatrix}e^{-10t}+\dots

    Anyone can chip in, if they'd like...I have a feeling there isn't another eigenvector...

    Does this make sense?

    --Chris
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by Chris L T521 View Post

    The eigenvector for \lambda = -10 would be found by solving the equation \begin{bmatrix}3 & 0 & 6 \\ 0 & 15 & 0 \\ 6 & {\color{red}-5} & 12\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bm  atrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
    Maybe I'm just with stupid, but why did you put that red -5 ?
    Though it doesn't change the result.



    \begin{bmatrix}-12 & 0 & 6 \\ 0 & 0 & 0 \\ 6 & {\color{red}0} & -3\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bma  trix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
    We have then the equations

    \left\{\begin{array}{ll} -12 v_1+6v_3=0 \\ 6v_1-3v_3=0 \end{array} \right.

    Okay, solve calmly these equations, you get :
    2v_1=v_3
    So you can take v_1=1 and v_3=2


    Now the difference is in v_2.
    I don't really remember why, but you have to take v_2=1 and v_2=0, because of matters of linear independence.
    If you think about it, it'd be logic. However, I'm sorry I don't know what the explanation is ><


    Letting v_2=0
    That was your mistake. What if v_2 \neq 0 ???
    I admit that was not really a mistake. Actually, it could be directly deduced that v_2=0 from your equations since 6v_1-3v_3=-3(2v_1-v_3)=0 and hence -5v_2=0.

    But as long as you made your equation wrong,....

    If you let something be a zero, then think about the alternative.
    Or is it just me who misunderstood your "letting" ?
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