My Math Forum Matrice - Eigenvalue and its eigenvectors

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October 20th, 2015, 09:15 AM   #1
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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 A-lambda matrix

Anybody who can figure this out?
Attached Images
 matrice.png (2.4 KB, 1 views)

 October 22nd, 2015, 01:59 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 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 Eigen-space?

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