
#1
Aug1211, 06:24 AM


#2
Aug1211, 07:15 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898

Your first eigenvector corresponding to [itex]\lambda= 3[/itex] is correct.
However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote [itex](A \lambda I)x= 0[/itex]: [tex]\begin{bmatrix}2 & 0 \\ 8 & 2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex] which gives [tex]\begin{bmatrix}2x_1 \\ 8x_1 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex] which then requires that [itex]2x_1= 0[/itex] and [itex]8x_1 2x_2= 0[/itex]. From the second [itex]x_2= 4x_1[/itex] but the first says [itex]x_1= 0[/itex] so [itex]x_2= 4(0)= 0[/itex]. There is NO nontrivial vector satisfying this. 1 is NOT an eigenvalue (I have no idea where you got "[itex]x_1+ x_2= 0[/itex]".) The characteristic equation is given by [tex]\left\begin{array}{cc}3 \lambda & 0 \\ 8 & 1 \lambda\end{array}\right= 0[/tex] Which is, of course, simply [itex](3\lambda)(1 \lambda)= 0[/itex]. 



#3
Aug1211, 03:48 PM


#4
Aug1211, 07:05 PM

P: 90

Please check my Eigenvector solutions.
If you compute what you think are an eigenvector and eigenvalue pair, stick them back in! They had better satisfy [itex] Ax = \lambda x[/itex] since that is, after all, the equation whose solutions you were looking for in the first place.




#5
Aug1211, 08:48 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898





#6
Aug1311, 03:24 AM

P: 15

Thanks Stringy and HallsofIvy :)



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