Question

I confuse when finding the eigenvector from eigenvalue . my matrix is (top: 3,0,-5; middle: 1/5, -1,0; bottom is1,1,-2) I got eigenvalue 0, and +- sqr(2) . from 0 I have the matrix lamdaI - a is (top -3,0,5; middle -1/5,1,0; bottom: -1,-1,2) and ...done

Answer 1

\[\left[\begin{matrix}-3 & 0 & 5 \\ \frac{ -1 }{ 5 } & 1 & 0\\-1 & -1 & 2\end{matrix}\right]\] That's the lamdaI -A

Answer 2

how to step up?

Answer 3

Is this the given matrix A?
\[\left[\begin{matrix}3 & 0 & -5\\ \frac{1}{5} & -1& 0 \\ 1 & 1 &-2\end{matrix}\right]\]

Answer 4

yes,

Answer 5

my discrete teacher allows me to get A - lamda I, but my linear teacher forces me apply lamdaI -A. so, sometimes, I confuse

Answer 6

My linear teacher taught me to find eigenvalues by solving |A-λI|=0...

Answer 7

I know. me too. the most difficulty is I must adjust to every teacher, come up with their way only. switch everyday.

Answer 8

\[\left|\begin{matrix}
3-\lambda&0&-5\\{1\over5}&-1-\lambda&0\\1&1&-2-\lambda
\end{matrix}\right|=0\\
(3-\lambda)(-1-\lambda)(-2-\lambda)-5\left[{1\over5}-(-1-\lambda)\right]=0\\
(3-\lambda)(1+\lambda)(2+\lambda)-1-5(1+\lambda)=0
\]

Answer 9

I got characteristic equation is \[\lambda^3 -2\lambda=0\]

Answer 10

I use the formula of L ^3 -trace (a)L^2 + (A11 +A22+A33) L - det (A) =0

Answer 11

Then solve A-λI =0 to find the corresponding eigenvector.

Answer 12

sure, and I got L =0 and L =+- sqr(2) . those are the right answer from my book. as I stated, I got problem on switch eigenvalue to eigenvector.only. not the above part

Answer 13

Did you get the eigenvalues by solving |A-λI| =0? or |λI-A| =0?

Answer 14

no. the book asks for eigenvalue on odd question, so I know I got the right one, but the eigenvector is in the even question, so I don't know

Answer 15

sorry, wrong attachment

Answer 16

eigenvector-> there exists a vector "v" for every non-zero eigen value..

Answer 17

but if you set:
\[Av=0,\\
-3x+5z=0\\
x-5y=0\\
x+y-2z=0
\]

Answer 18

When lambda=0,
\[\left[\begin{matrix}3-0 & 0 & -5\\ \frac{1}{5} & -1-0& 0 \\ 1 & 1 &-2-0\end{matrix}\right] \rightarrow \left[\begin{matrix}3 & 0 & -5\\ \frac{1}{5} & -1& 0 \\ 1 & 1 &-2\end{matrix}\right] \rightarrow \left[\begin{matrix}1 & 1 &-2\\ \frac{1}{5} & -1& 0 \\ 3 & 0 & -5\end{matrix}\right]\]\[\rightarrow \left[\begin{matrix}1 & 1 &-2\\ 0& -\frac{6}{5}& \frac{2}{5} \\ 0 & -3 & 1\end{matrix}\right]\rightarrow \left[\begin{matrix}1 & 1 &-2\\ 0&3& -1 \\ 0 & 0 & 0\end{matrix}\right]\]
So, the eigenvector corresponding to the eigenvalue lamda =0 is\[v_1=\left(\begin{matrix}5 \\ 1\\3\end{matrix}\right)\]

Answer 19

1) x=(5/3)z
2) x=5(y)
so, if x=5, y=1 and z=3
so the first vector is [5,1,3]

Answer 20

like Callisto

Answer 21

And you need to solve the other two. Just the same method to solve it, that is to solve
A - λI =0

Answer 22

thanks guys. I got it now.

Answer 23

can I close this post? everything done, isn't it?

Answer 24

You can, if you want!

Answer 25

thnks let the space for anyone who need help.