How to find the eigenvector and determine type, special case.

[Sheng Zang]

I got a question like this,
As I solved, I find the critical point is (0,0) and A= (0                 1)
                                                                          (-6x^2+8     0)
after plug(0,0) into, it becomes (0, 1)
                                              (0, 0)
so, λ1=0=λ2, how to solve eigenvector and determine the type of critical point?

Thanks for help.

[Victor Ivrii]

your post is incomprehensible

How to find eigenvector? See Linear algebra course and Sections 7.1, 7.2 of our textbook

[Sheng Zang]

I mean forλ1=0, matrix is A=(0 1)  *(x1) =(0)
                                           (0 0)   (y1)    (0)
the eigenvector is (1,0) which is horizontal.

and since λ is repeated, i.e λ1=λ2=0,
we solve (0 1)  *(x2) =(1)
              (0 0)   (y2)    (0)
so, another eigenvector is (0,1), which is vertical.

Am I correct?

[Victor Ivrii]

Except $(0,1)$ is not an eigenvector but a generalized e.v.

[Sheng Zang]

Also what is the type of critical point?  proper or improper node? And how to draw pic?( I don't think this is as usual as we draw (im)proper node picture)

[Victor Ivrii]

Improper node. See in Textbook. Forum is not a substitution for it