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Thread: 5.5.17 Find the eigenvalues of the matrix

  1. #1
    Super Member bigwave's Avatar
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    Cool 5.5.17 Find the eigenvalues of the matrix

    $\textsf{Find the eigenvalues of the matrix:}$
    $$\left[\begin{array}{rrr} 0& 0& 0\\ 0&3& 4\\ 0& 0&-2
    \end{array} \right]=
    \left[\begin{array}{rrr}
    \lambda & 0 &0\\
    0 &\lambda &0\\
    0 &0 &\lambda
    \end{array}\right]$$
    $\textsf{so $A-\lambda x =0$ then}$
    $$\left[\begin{array}{rrr}
    (0-\lambda)& (0-0)& (0-0)\\
    (0-0)&(3-\lambda)& (0-4)\\
    (0-0)& (0-0)&(-2-\lambda)
    \end{array}\right]=0$$
    $\textsf{so then} $
    $$(0-\lambda)(3-\lambda)(-2-\lambda)=0$$
    $\textsf{therefore}$
    $$\lambda =0,\quad\lambda =-3,\quad\lambda =2$$


    ok this was the bk answer but not sure if my process was right?
    Last edited by bigwave; Mar 29th 2018 at 01:11 PM. Reason: extra matrix
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  2. #2
    MHF Contributor

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    Re: 5.5.17 Find the eigenvalues of the matrix

    Yes, given the matrix $\displaystyle A= \begin{bmatrix}0 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -2\end{bmatrix}$ then the
    "eigenvalue" equation is $\displaystyle |A- \lambda I|= \left|\begin{array}{ccc}-\lambda & 0 & 0 \\ 0 & 3- \lambda & 4 \\ 0 & 0 & -2- \lambda \end{array}\right|= 0$. Expanding the determinant on the first row that is $\displaystyle -\lambda \left|\begin{array}{cc}3- \lambda & 4 \\ 0 & -2- \lambda \end{array}\right|= (-\lambda(3- \lambda)(-2- \lambda)= 0$ so that the eigenvalues are 0, -2, and 3. You can check that by looking for corresponding eigenvectors.

    $\displaystyle \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ is an eigenvector corresponding to eigenvalue 0 if and only if $\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -2 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= 0\begin{bmatrix}x \\ y \\ z\end{bmatrix}$. That reduces to $\displaystyle \begin{bmatrix}0 \\ 3y+ 4z \\ -2z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ so that we must have 3y+ 4z= 0 and -2z= 0. Those equations give y= z= 0 but there is no condition on x. $\displaystyle \begin{bmatrix}x \\ 0 \\ 0 \end{bmatrix}$ is an eigenvector for any value of x (generally, if v is a eigenvector corresponding to a given eigenvalue, so is any multiple of v).

    Similarly, if $\displaystyle \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ is an eigenvector corresponding to eigenvalue -2 then $\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -2 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= -2\begin{bmatrix}x \\ y \\ z\end{bmatrix}$. That reduces to $\displaystyle \begin{bmatrix}0 \\ 3y+ 4z \\ -2z\end{bmatrix}= \begin{bmatrix}-2x \\ -2y \\ -2z\end{bmatrix}$ so we must have 0= -2x, 3y+ 4z= -2y, and -2z= -2z. The first equation gives x= 0, the second gives 4z= -5y and the third is true for all z. If, in 4z= -5y, we take y= 4, we have z= -5. So $\displaystyle \begin{bmatrix}0 \\ 4 \\ -5\end{bmatrix}$ and any multiple of that is an eigenvector corresponding to eigenvalue -2.

    Finally, if $\displaystyle \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ is an eigenvector corresponding to eigenvalue 3 then $\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -2 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= 3\begin{bmatrix}x \\ y \\ z\end{bmatrix}$. That reduces to $\displaystyle \begin{bmatrix}0 \\ 3y+ 4z \\ -2z\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$ so we must have 0= 3x, 3y+ 4z= 3y, and -2z= 3z. The first and third equations reduce to x= z= 0 and them the third equation is 3y= 3y which is true for all y. Eigenvectors corresponding to eigenvalue 3 is of the form $\displaystyle \begin{bmatrix} 0 \\ y \\ 0 \end{bmatrix}$.

    (If we had tried a number that was NOT an eigenvalue the three equations would have reduced to x= y= z= 0.)
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