I don't know what are you doing. For the corresponding eigenspace is:
and a basis is:
Fernando Revilla
So I've to find the eigenvalues and a basis for each corresponding eigenspaces.
Matrix A=[(2,0,0), (1,-1,0), (-1,3,3)]
After finding the determinant, I got(λ-2) (λ+1) (λ-3)=0. So λ = 2, -1, 3 are the eigenvalues.
[(λ-2, 0,0),(-1, λ+1, 0),(1,-3,λ-3)]
For λ=2; -v_2+v_3=0
3v_2-3v_3=0
-v_3=0
Let v_3 = t, then v_2=-t, but what is v_1? Is it 0? So the span is [0, -1, 1]?
For λ=-1; -3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0
Let v_3=t, then what is v_2 and v_1? Are they both 0? So the span is [0,0,1]?
For λ=3; v_1-v_2+v_3=0
4v_2-3v_3=0
Let v_3=t, then v_2=(3/4)t and v_1=(-1/4)t. So the span is [(-1/4), (3/4), 1]? I think I got this right.
I don't know what are you doing. For the corresponding eigenspace is:
and a basis is:
Fernando Revilla
More than that. Check:
Fernando Revilla
Yes, for any upper or lower triangular matrices, the eigenvalues are the number on the main diagonal.
Okay. I tend to think of eigenvector [itex]\lambda[/itex] and eigenvector v as satifying . You are writing it is which is the same thing.[(λ-2, 0,0),(-1, λ+1, 0),(1,-3,λ-3)]
What happened to ? The equations, with areFor λ=2; -v_2+v_3=0
3v_2-3v_3=0
-v_3=0[/tex]
The first equation does NOT say " ", it is true for any .
The second equation says and the third that
Taking , an eigenvector, corresponding to eivenvalue 2, is [3, 1, 0] and the eigenspace is spanned by [3, 1, 0].
(Added: If those were to be columns rather then rows, then your equations are
The third equation gives , both first and second equations give but there is no in any of the equations so it can be anything: [t, 0, 0]= t[1, 0, 0]. The eigenspace is spanned by [1, 0, 0])
Of course, it is easy to check whether or not a given vector is an eigenvector. Did you do that?
Let v_3 = t, then v_2=-t, but what is v_1? Is it 0? So the span is [0, -1, 1]?
For λ=-1; -3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0
Let v_3=t, then what is v_2 and v_1? Are they both 0? So the span is [0,0,1]?
For λ=3; v_1-v_2+v_3=0
4v_2-3v_3=0
Let v_3=t, then v_2=(3/4)t and v_1=(-1/4)t. So the span is [(-1/4), (3/4), 1]? I think I got this right.
To avoid cofusion, I suggest you next time to write in code. Alternatively, use (a,b,...)^T for columns.
Fernando Revilla
Impossible. If is an eigenspace then .
-3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0
Fernando Revilla