## liliy Group Title Find all the eigenvalues for the given matrices. if it is a 3x3. and a polynomial. how do i know what to break it down into? 2 years ago 2 years ago

1. liliy Group Title

[ 4 0 1 -1 -6 -2 5 0 0 ] $\lambda ^3+2 lambda ^2 - 29 lambda -30]/ 2. UnkleRhaukus Group Title \[\textbf Av=\lambda v$ $\left(\textbf A-\lambda \textbf I\right)v =0$ $\left|\textbf A-\lambda \textbf I\right|v =0$ $\left|\left(\begin{array}\ 4&0&1\\-1&-6&-2\\5&0&0\end{array}\right)-\left(\begin{array}\ \lambda&0&0\\0&\lambda & 0\\0 &0 &\lambda \end{array}\right)\right|=0$ $\left|\left(\begin{array}\ 4-\lambda&0&1\\-1&-6-\lambda&-2\\5&0&0-\lambda\end{array}\right)\right|=0$ $(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]-0[-1\times(0-\lambda)]+1[-1\times0-(-2\times5)]=0$

3. UnkleRhaukus Group Title

*$(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]$$\qquad-0[-1\times(0-\lambda)-(5\times-2)]$$\qquad\qquad+1[-1\times0-(-6-\lambda)]=0$

4. liliy Group Title

right, i know how to do all the calculations but when you get the the last step: when its all combined how do you know what lamda itself is equal to? ( like, you look at the constant and see its factors or something and do something..idk)_)

5. UnkleRhaukus Group Title

well at the last spet you'll have a bunch of factors =0 so to get zero at least one of the factors is zero

6. UnkleRhaukus Group Title

something like this only i have made mistakes $(\lambda-4)[(6+\lambda)\lambda]+[6+\lambda]=0$$(6\lambda+\lambda^2-24-4\lambda)\lambda+6+\lambda=0$$\lambda^3+2\lambda^2-24\lambda+\lambda+6=0$$\lambda^3+3\lambda^2-24\lambda+6=0$$(\lambda-a)(\lambda-b)(\lambda-c)=0$ lambda would be a,b,c

7. liliy Group Title

but how do you know to break it into a b c?

8. UnkleRhaukus Group Title

how to factorize? you could try polynomial long division

9. liliy Group Title

ugh

10. liliy Group Title

$\lambda^3 +2\lambda^2-29\lambda-30$

11. liliy Group Title

how would you go about doing this? possibilities: +/- : 1,2,3,5,6,10,30,15

12. liliy Group Title

if you plug in -1, you get =0.then what?

13. UnkleRhaukus Group Title

your expression does factorizes nicely, try dividing by $$(\lambda+1)$$

14. liliy Group Title

i dont know how... i knwo that sounds dumb but how do u do that?

15. UnkleRhaukus Group Title

|dw:1341885937030:dw|

16. UnkleRhaukus Group Title

|dw:1341885983310:dw|

17. UnkleRhaukus Group Title

so$\lambda^3 +2\lambda^2-29\lambda-30=(\lambda+1)(\lambda^2+\lambda-30)$

18. UnkleRhaukus Group Title

now just factorize the quadratic factor...

19. liliy Group Title

thanks. also: do you know about diagnolization?