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October 20th, 2015, 09:15 AM  #1 
Newbie Joined: Oct 2015 From: Poland Posts: 1 Thanks: 0  Matrice  Eigenvalue and its eigenvectors
I have a matrix  given in the picture I uploaded. They give the information that 2(1.5+0.5a) is an eigenvalue for the vector A. I now need to find the eigenvectors for that exact eigenvalue. I can't really figure it out, but i thought about using this: (A−λE)x = 0 But I'm getting some complicated shit because I don't have the values for the eigenvector, I'm just calling the vector <v1,v2,v3> and multiply it onto the Alambda matrix Anybody who can figure this out? 
October 22nd, 2015, 01:59 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
Tedious but straight forward. Given that 2(1.5+ 0.5a)= 3+a is an Eigen value then $\displaystyle A \lambda E= \begin{bmatrix} 1+ a & 1+ a & 1+ a \\ 1+ a & 1+ a & 1+ a \\3(1+ a) & 3(1+ a) & 3(1+ a)\end{bmatrix}$. So $\displaystyle (A \lambda E)(v_1, v_2, v_3)= $$\displaystyle \begin{bmatrix}(1+ a)(v_1+ v_2+ v3) \\ (1+ a)(v_1+ v_2+ v_3) \\ 3(1+ a)(v_1+ v_2+ v_3)\end{bmatrix}$$\displaystyle = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$. So we must have $\displaystyle (1+ a)(v_1+ v_2+ v_3)= 0$, $\displaystyle (1+ a)(v_1+ v_2+ v_3)= 0, 3(1+ a)(v_1+ v_2+ v3)= 0$. If a= 1, so that 1+ a= 0, that is true for all vectors. If a is not 1, then $\displaystyle v_1+ v_2+ v_3= 0$. We can solve for any one of those in terms of the other two. For example $\displaystyle v_3= v_1 v_2$ so we can write $\displaystyle (v_1, v_2, v_3)= (v_1, v_2, v_1 v_2)= (v_1, 0, v_1)+ (0, v_2, v_2)= v_1(1, 0, 1)+ v_2(0, 1, 1)$. Do you see a basis for this Eigenspace? 

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eigenvalue, eigenvectors, matrice 
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