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Math Help - Finding eigenvectors

  1. #1
    Super Member Showcase_22's Avatar
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    Finding eigenvectors

    Find the eigenvectors of the matrix \begin{pmatrix}<br />
{-3}&{17}&{-4}\\<br />
{-2}&{9}&{-2}\\<br />
{-2}&{8}&{-1}\\<br />
\end{pmatrix}

    I've already found the eigenvalues which come out to be 1, 2+i and 2-i.

    To find the eigenvectors I have to find v such that (A-\lambda I)v=0.

    I already found the eigenvector for \lambda=1. I'm having trouble with it for the complex cases.

    Here's my working for \lambda=2-i:

    A-(2-i)I=\begin{pmatrix}<br />
{-5+i}&{17}&{-4}\\<br />
{-2}&{7+i}&{-2}\\<br />
{-2}&{8}&{i-3}\\<br />
\end{pmatrix}

    Send R_3=R_3-R_2

    A-(2-i)I=\begin{pmatrix}<br />
{-5+i}&{17}&{-4}\\<br />
{-2}&{7+i}&{-2}\\<br />
{0}&{1-i}&{i-1}\\<br />
\end{pmatrix}

    Send R_1=R_1-2R_2

    A-(2-i)I=\begin{pmatrix}<br />
{-1+i}&{3-2i}&{0}\\<br />
{-2}&{7+i}&{-2}\\<br />
{0}&{1-i}&{i-1}\\<br />
\end{pmatrix}


    (A-(2-i)I)v=\begin{pmatrix}<br />
{-1+i}&{3-2i}&{0}\\<br />
{-2}&{7+i}&{-2}\\<br />
{0}&{1-i}&{i-1}\\<br />
\end{pmatrix} \begin{pmatrix}<br />
{v_1}\\<br />
{v_2}\\<br />
{v_3}\\<br />
\end{pmatrix}=0

    (-1+i)v_1+(3-2i)v_2=0 \Rightarrow -v_1+3v_2+i(v_1-2v_2)=0 \Rightarrow v_1=3v_2 \ , \ v_1=2v_2

    (1-i)v_2+(i-1)v_3=0 \Rightarrow v_2-v_3+i(-v_2+v_3) \Rightarrow v_2=v_3

    The only vector that satisfies these conditions is the zero matrix which cannot be an eigenvector by definition!
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Find the eigenvectors of the matrix \begin{pmatrix}<br />
{-3}&{17}&{-4}\\<br />
{-2}&{9}&{-2}\\<br />
{-2}&{8}&{-1}\\<br />
\end{pmatrix}

    I've already found the eigenvalues which come out to be 1, 2+i and 2-i.

    To find the eigenvectors I have to find v such that (A-\lambda I)v=0.

    I already found the eigenvector for \lambda=1. I'm having trouble with it for the complex cases.

    Here's my working for \lambda=2-i:

    A-(2-i)I=\begin{pmatrix}<br />
{-5+i}&{17}&{-4}\\<br />
{-2}&{7+i}&{-2}\\<br />
{-2}&{8}&{i-3}\\<br />
\end{pmatrix}

    Send R_3=R_3-R_2

    A-(2-i)I=\begin{pmatrix}<br />
{-5+i}&{17}&{-4}\\<br />
{-2}&{7+i}&{-2}\\<br />
{0}&{1-i}&{i-1}\\<br />
\end{pmatrix}

    Send R_1=R_1-2R_2

    A-(2-i)I=\begin{pmatrix}<br />
{-1+i}&{3-2i}&{0}\\<br />
{-2}&{7+i}&{-2}\\<br />
{0}&{1-i}&{i-1}\\<br />
\end{pmatrix}


    (A-(2-i)I)v=\begin{pmatrix}<br />
{-1+i}&{3-2i}&{0}\\<br />
{-2}&{7+i}&{-2}\\<br />
{0}&{1-i}&{i-1}\\<br />
\end{pmatrix} \begin{pmatrix}<br />
{v_1}\\<br />
{v_2}\\<br />
{v_3}\\<br />
\end{pmatrix}=0

    (-1+i)v_1+(3-2i)v_2=0 \Rightarrow -v_1+3v_2+i(v_1-2v_2)=0 \Rightarrow v_1=3v_2 \ , \ v_1=2v_2

    (1-i)v_2+(i-1)v_3=0 \Rightarrow v_2-v_3+i(-v_2+v_3) \Rightarrow v_2=v_3

    The only vector that satisfies these conditions is the zero matrix which cannot be an eigenvector by definition!
    Hi - Are you assuming v1,v2,v3 to be reals. I guess that is the error.
    They can be complex.

    I am not 100% sure, but somewhat inclined towards the above so shot this post.
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  3. #3
    Super Member Showcase_22's Avatar
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    Let v_1=a+bi, v_2=c+di and v_3=e+fi.

    If a=1, b=0 then  -1+i+(3-2i)(c+di)=0 \Rightarrow -1+i+3c+3di-2ci+2d=0 \Rightarrow -1+3c+2d+i(1+3d-2c)=0.

    \Rightarrow 3c+2d=1 and 3d-2c=-1.

    \Rightarrow 2d-\frac{4}{3}c=-\frac{2}{3}

    3c+\frac{4}{3}c-\frac{2}{3}=1 \Rightarrow \frac{13}{3}c=\frac{5}{3} \Rightarrow 13c=5 \Rightarrow c=\frac{5}{13}

    This gives d=\frac{2c-1}{3}=\frac{\frac{10}{13}-1}{3}=-\frac{1}{13}

    So v_2=\frac{1}{13}(5-i)

    Similarly,

    (1-i)\left( \frac{1}{13}(5-i) \right)+(i-1)(e+fi)=0

    \frac{1}{13}(1-i)(5-i)+ei-e-f-fi=0

    \frac{1}{13}(5-5i-i-1)+ei-e-f-fi=0

    \frac{5}{13}-\frac{5}{13}i-\frac{1}{13}i-\frac{1}{13}+ei-e-f-fi=0

    \frac{4}{13}-e-f+  i \left(\frac{4}{13}+e-f \right)=0

    \Rightarrow e+f=\frac{4}{13} and f-e=\frac{4}{13}

    2f=\frac{8}{13} \Rightarrow f=\frac{4}{13}

    e=0

    So the eigenvector corresponding to \lambda=2-i is u_1=\begin{pmatrix}<br />
{1}\\<br />
{\frac{1}{13}(5-i)}\\<br />
{\frac{4}{13}i}\\<br />
\end{pmatrix}

    Does this look about right?
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  4. #4
    Super Member
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    Quote Originally Posted by Showcase_22 View Post
    Let v_1=a+bi, v_2=c+di and v_3=e+fi.

    If a=1, b=0 then  -1+i+(3-2i)(c+di)=0 \Rightarrow -1+i+3c+3di-2ci+2d=0 \Rightarrow -1+3c+2d+i(1+3d-2c)=0.

    \Rightarrow 3c+2d=1 and 3d-2c=-1.

    \Rightarrow 2d-\frac{4}{3}c=-\frac{2}{3}

    3c+\frac{4}{3}c-\frac{2}{3}=1 \Rightarrow \frac{13}{3}c=\frac{5}{3} \Rightarrow 13c=5 \Rightarrow c=\frac{5}{13}

    This gives d=\frac{2c-1}{3}=\frac{\frac{10}{13}-1}{3}=-\frac{1}{13}

    So v_2=\frac{1}{13}(5-i)

    Similarly,

    (1-i)\left( \frac{1}{13}(5-i) \right)+(i-1)(e+fi)=0

    \frac{1}{13}(1-i)(5-i)+ei-e-f-fi=0

    \frac{1}{13}(5-5i-i-1)+ei-e-f-fi=0

    \frac{5}{13}-\frac{5}{13}i-\frac{1}{13}i-\frac{1}{13}+ei-e-f-fi=0

    \frac{4}{13}-e-f+  i \left(\frac{4}{13}+e-f \right)=0

    \Rightarrow e+f=\frac{4}{13} and f-e=\frac{4}{13}

    2f=\frac{8}{13} \Rightarrow f=\frac{4}{13}

    e=0

    So the eigenvector corresponding to \lambda=2-i is u_1=\begin{pmatrix}<br />
{1}\\<br />
{\frac{1}{13}(5-i)}\\<br />
{\frac{4}{13}i}\\<br />
\end{pmatrix}

    Does this look about right?
    Seems right to me. But I am like not too proficient with the subject. Maybe you would want to back substitute and check once.
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  5. #5
    MHF Contributor

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    By the way, are you working with a vector space over the real numbers or the complex numbers. Since the matrix you give has only real entries, it could be either one. But if you are working with a vector space over the real numbers, then only "1" is an eigenvalue.
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  6. #6
    Super Member Showcase_22's Avatar
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    The question itself doesn't say, but i'm guessing it's over the complex numbers.

    Also, I think there's something wrong with my above working. Checking on octave gives:

    A =

    -3 17 -4
    -2 9 -2
    -2 8 -1

    P =

    1.00000 - 0.00000i
    0.38462 + 0.07692i
    0.00000 - 0.30769i

    octave-3.0.1:4> B=A*P
    B =

    3.5385 + 2.5385i
    1.4615 + 1.3077i
    1.0769 + 0.9231i

    octave-3.0.1:6> B./P
    ans =

    3.5385 + 2.5385i
    4.3077 + 2.5385i
    -3.0000 + 3.5000i

    But B./P should be an eigenvector, either 2-i or 2+i.

    Furthermore, actually computing the answer on matlab gives:

    A =

    -3 17 -4
    -2 9 -2
    -2 8 -1

    octave-3.0.1:2> [V,D] = eig(A)
    V =

    0.87447 + 0.00000i 0.87447 - 0.00000i -0.70711 + 0.00000i
    0.33634 + 0.06727i 0.33634 - 0.06727i -0.00000 + 0.00000i
    0.33634 + 0.06727i 0.33634 - 0.06727i 0.70711 + 0.00000i

    D =

    2.00000 + 1.00000i 0.00000 + 0.00000i 0.00000 + 0.00000i
    0.00000 + 0.00000i 2.00000 - 1.00000i 0.00000 + 0.00000i
    0.00000 + 0.00000i 0.00000 + 0.00000i 1.00000 + 0.00000i

    ...and this is clearly different to what I have.

    From what I can see, my initial method to find the eigenvectors should have worked!
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