Actually that gives you A - [itex]\lambda[/itex]I, which is different from A (unless [itex]\lambda[/itex] happens to be zero).So you sub lambda into your matrix. This gives you a matrix A.
I know what you mean, but what you really want is to find a solution of (A - [itex]\lambda[/itex]I)x = 0.Now you need to find a vector x such that Ax=0.
This involves solving a system of equations. The solutions of that system should give you the eigenvectors...
Assuming this is A - [itex]\lambda[/itex]I, yes, <-1/3, 1> is an eigenvector, and so are <-1, 3>, <1/3, -1>, <1, -3>, and many more. You can easily check whether a vector x is an eigenvector with associated eigenvalue [itex]\lambda[/itex] by verifying that Ax = [itex]\lambda[/itex]x, or equivalently, that (A - [itex]\lambda[/itex]I)x = 0.Ah rite I see so when lambda = 0 eigenvector is (1,-1)? Its just confusing as when matrix wa:
3,1
0,0
eigenvector is (-1/3,1) ???