For lamda = -1: its (A+I)*v=0 simplifies to:

[1 3/2 | 0 ]

[0 0 | 0 ] so converting this into vector form

[v1] v2*[-3/2]

[v2] = [ 1 ] So v2=t then the eigenvector is

[-3] * t for t belong to R

[2]

I understand for the above step it was taken from the linear simultaneous equation system:

1) v1+(3/2)v2=0

2) 0(v1) + 0(v2) = 0

Because of the 2nd row being all 0s, we have an infinite amount of solutions, which is why we had to add a "t belong to R" parameter to the eigenvector.

BUT for lamda = 4:

{ [1 3] - 4* [1 0] } * v = 0

{ [2 2] [0 1] }

it simplifies down to

[1 0 | 0]

[0 1 | 0]

If I was to follow the same method of linear equation system as above, how is something like this converted into the eigenvector

[1] * t for t belong to R?

[1]

since: v1+0(v2) = 0

0(v1)+v2 = 0 shouldn't v =

[0] instead of [1]

[0] instead of [1] ? And that it is a unique solution without even the need for parameter t?