I obtained eigenvalues of 0,-0.9 and 1 from the matrix:

|0.1 1 1|

|0.5 0 0|

|0.4 0 0|

Now I need to find the eigenvectors. I'm thinking I set one of X1 X2 or X3 to the variable t and another s to obtain the eigenvectors. Any help?

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- Mar 1st 2009, 01:35 PMthehollow89Find eigenvectors from given eigenvalues
I obtained eigenvalues of 0,-0.9 and 1 from the matrix:

|0.1 1 1|

|0.5 0 0|

|0.4 0 0|

Now I need to find the eigenvectors. I'm thinking I set one of X1 X2 or X3 to the variable t and another s to obtain the eigenvectors. Any help? - Mar 1st 2009, 02:12 PMHallsofIvy
Yes- use the definition of "eigenvector" If <x y z> is an eigenvector corresponding to eigenvalue 0 then we must have

$\displaystyle \begin{bmatrix}0.1 & 1 & 1 \\ 0.5 & 0 & 0 \\0.4 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

which is the same as the three equations 0.1x+ y+ z= 0, 0.5x= 0, 0.4 x= 0. The last two equations tell us that x must be 0. Putting that into the first equation, y+ z= 0 or z= -y. An eigenvector corresponding to eigenvalue 0 is of the form <0 y - y> or y<0 1 -1>.

If <x y z> is an eigenvector corresponding to eigenvalue -.9 just replace the "0"s on the right by -.9x, -.9y, -.9z: 0.1x+ y+ z= -0.9x, 0.5x= -0.9y, 0.4x= -0.9z. The last two equations can be written as y= (-0.5/0.9)x and z= (-0.4/0.9)x. Putting those into the first equation gives 0.1x- (0.5/0.9)x- (0.4/0.9)x= -0.9 x which is true no matter what x is. That serves as a check that -0.9**is**an eigenvalue. But from y= (-5/9)x and z= (-4/9)x we can write the eigenvector as <x, (-5/9)x, (-4/9)x>= x<1, -5/9, -4/9> or even (x/9)<9, -5, -4>.

I'll let you do eigenvalue 1.