Eigenvectors and steady state vectors

[jellybone]

I'm doing a task from my book about population dynamics where i have to set up a transition matrix, find eigenvalues, egenvectors and then a steady state vector.
I've tried doing some of it and I'm not sure if It's quite right, but some guidance in the right direction would be appreciated.

Here are the values extracted from the task into a transition matrix.
Transition matrix:

0.8	0	0
0	1.2	0
0	0	1

Found the Eigenvalues, 0,8, 1,2 and 1.


Eigenvectors:
for eigenvalue 0.8

Setting it up in a linear system:
0.8x + 0y + 0z = 0.8x
0x + 1.2y + 0z = 0.8y
0x + 0y + z = 0.8z

This brings me to x=s, y=0.4s and z = 0.2s.

So the vector will look like this:
[ 1 ]
s [0.4]
[0.2]

When i calculate the eigenvector for eigenvalue 0.8 on a calculator online, i get the vector:
[1]
[0]
[0]

Why is this different from my answer?

 

[Ishuda]

I'm doing a task from my book about population dynamics where i have to set up a transition matrix, find eigenvalues, egenvectors and then a steady state vector.
I've tried doing some of it and I'm not sure if It's quite right, but some guidance in the right direction would be appreciated.

Here are the values extracted from the task into a transition matrix.
Transition matrix:

0.8	0	0
0	1.2	0
0	0	1

Found the Eigenvalues, 0,8, 1,2 and 1.


Eigenvectors:
for eigenvalue 0.8

Setting it up in a linear system:
0.8x + 0y + 0z = 0.8x
0x + 1.2y + 0z = 0.8y
0x + 0y + z = 0.8z

This brings me to x=s, y=0.4s and z = 0.2z.

...

First fix the typos in you matrix equation results "This brings me to x=x, y=(2/3)y and z = 0.8 z." That is
A v = \(\displaystyle \begin{pmatrix} 0.8& 0.0& 0.0\\ 0.0& 1.2& 0.0 \\ 0.0& 0.0& 1.0 \end{pmatrix}
\begin{pmatrix} x\\ y\\ z\end{pmatrix}\, =\,
\begin{pmatrix} 0.8x\\ 1.2y\\ 1.0z\end{pmatrix}\) = 0.8 v = \(\displaystyle \begin{pmatrix} 0.8x\\ 0.8y\\ 0.8z\end{pmatrix}\)

The solution to that is x=x, y=0, and z=0. So for any x not equal to zero and v=(x,0,0)^T, we have Av=0.8v. Choosing x=1 gives the solution on the online calculator. Note that this implies that if v is an eigenvector and a is any non-zero constant, then av is an eigenvector.