Question

x'=Ax where \[A=\left[\begin{matrix}-3&0&2\\1&-1&0\\-2&-1&0\end{matrix}\right]\] I got characteristic equation \[\lambda ^3+4\lambda^2+7\lambda+6=0\\\lambda_1=-2;~\lambda_{2,3}=-1\pm\sqrt{2}i\] continue on comment

Answer 1

for \(\lambda_1 =-2\) I got the eigenvector \(\left(\begin{matrix}2\\-2\\1\end{matrix}\right)\)
for \(\lambda_2=-1+\sqrt{2}i\) I got the matrix \[\left[\begin{matrix}-2-\sqrt{2}i&0&2\\1&\sqrt{2}i&0\\-2&-1&1+\sqrt{2}i\end{matrix}\right]\]but don't know how to get the eigenvector from this. Please, help

Answer 2

I step away for awhile, when you stop by, please leave your help. Thanks in advance

Answer 3

just had to check to make sure your lambdas was good:

-2, \(-1\pm i\sqrt2\) are fine

Answer 4

the next step is to just plug in the lambdas and row reduce to echelon form to define the eugene spaces i believe

Answer 5

i might have to review that vector step ....

Answer 6

lol, i used l=2 not -2 on my paper

Answer 7

1 0 -2 x0 = x3( 2)
0 1 2 x2 = x3(-2) so (2,-2,1) is an eugene vector .... good
0 0 0 x3 = x3( 1)

Answer 8

\[\left[\begin{matrix}
-3-(-1+i\sqrt2)&0&2\\
1&-1-(-1+i\sqrt2)&0\\
-2&-1&-(-1+i\sqrt2)\end{matrix}\right]\]

\[rref:\left[\begin{matrix}
-2-i\sqrt2&0&2\\
1&-i\sqrt2&0\\
-2&-1&1-i\sqrt2\end{matrix}\right]\]

Answer 9

\[\left[\begin{matrix}
-2-i\sqrt2&0&2\\
1&-i\sqrt2&0\\
-2&-1&1-i\sqrt2\end{matrix}\right]\]

\[\left[\begin{matrix}
1&-i\sqrt2&0\\
-2&-1&1-i\sqrt2\\
-2-i\sqrt2&0&2\\
\end{matrix}\right]\]

\[\left[\begin{matrix}
1&-i\sqrt2&0\\
0&-1-2i\sqrt2&1-i\sqrt2\\
0&2-2i\sqrt2&2\\
\end{matrix}\right]\]

\[\left[\begin{matrix}
1&-i\sqrt2&0\\
0&1&\frac{1-i\sqrt2}{-1-2i\sqrt2}\\
0&1&\frac{2}{2-2i\sqrt2}\\
\end{matrix}\right]\]

\[\left[\begin{matrix}
1&0&i\sqrt2\frac{1-i\sqrt2}{-1-2i\sqrt2}\\
0&1&\frac{1-i\sqrt2}{-1-2i\sqrt2}\\
0&0&\frac{2}{2-2i\sqrt2}-\frac{1-i\sqrt2}{-1-2i\sqrt2}\\
\end{matrix}\right]\]

etc ...

Answer 10

might have to wolf it to be sure i didnt get a typing error in there

Answer 11

http://www.wolframalpha.com/input/?i=rref%7B%7B-3-%28-1%2Bi*sqrt2%29%2C0%2C2%7D%2C%7B1%2C-1-%28-1%2Bi*sqrt2%29%2C0%7D%2C%7B-2%2C-1%2C-%28-1%2Bi*sqrt2%29%7D%7D

Answer 12

it messier, but runniong some conjugates and such is just more time consuming then difficult perse

Answer 13

@amistre64 you mean I have to do the whole stuff to get the rref? WOW!!

Answer 14

lol, yes ... and you also have to do it correctly to boot

Answer 15

Omg, I thought there is a shortcut. hihihih.....

Answer 16

shortcut? in linear algebra? ... the only shortcut im aware of is a computer

Answer 17

hahahaaaaa. OK, if I have to, no choice for me. Thanks a lot

Answer 18

good luck ;)

Answer 19

thank you

Answer 20

let the eigenvector is \[\left(\begin{matrix}a\\b\\c\end{matrix}\right)\] from the matrix above , I have\[\left[\begin{matrix}
1&-i\sqrt2&0\\
-2&-1&1-i\sqrt2\\
-2-i\sqrt2&0&2\\
\end{matrix}\right]\rightarrow \left[\begin{matrix}
1&-i\sqrt2&0\\
0&-1-2\sqrt{2}i&1+\sqrt{2}i\\
1&0&\frac{1}{6}(2-\sqrt{2}i)\\
\end{matrix}\right]\]
To get it, I do : 2*row1 + row 2 to get row 2
and 1/6 *(2-sqrt(2) i) *row3 to get row 3
\[\text{My question is: can I step up by doing this}\]
\[a = i\sqrt{2}b\\a=\frac{1}{6}(2-\sqrt{2}i)c \]and my \[\left(\begin{matrix}a\\b\\c\end{matrix}\right)=\left(\begin{matrix}1\\\frac{-i\sqrt{2}}{2}\\2+\sqrt{2}i\end{matrix}\right)\]
@amistre64