# Thread: 5.5.17 Find the eigenvalues of the matrix

1. ## 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?

2. ## 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.)