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Vectors $(1,0,0),(0,1,1),(1,1,1)$ are eigenvectors of matrix $A$. Prove that vector $(1,2,2)$ is eigenvector of matrix $A$.

We have:

$$A(1,0,0) = \lambda_1 (1,0,0) \\ A(0,1,1) = \lambda_2 (0,1,1) \\ A(1,1,1) = \lambda_3 (1,1,1) $$

Furthermore I notice:

$$(1,1,1) = (1,0,0) + (0,1,1) \\ (1,2,2) = (1,0,0) + 2(0,1,1)$$

Because $A(1,1,1) = A(1,0,0)+A(0,1,1)$ we have $\lambda_3 (1,1,1) = \lambda_1 (1,0,0) + \lambda_2 (0,1,1)$ hence $\lambda_1 = \lambda_2 = \lambda_3$. So let $\lambda_1 = \lambda_2 = \lambda_3 = \lambda.$

Next I have $A(1,2,2) = A(1,0,0)+ 2A(0,1,1) = \lambda(1,0,0) + 2 \lambda(0,1,1) = \lambda (1,2,2)$.

Does is work?

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    $\begingroup$ Yes, that works. $\endgroup$ – Daniel Fischer Aug 10 '13 at 13:20
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Consider a more general case (and deliberately avoiding the numbers to make it more clear) $\mathbf{A}\in\mathbb{C}^{n\times n}$ has Eigen vectors $\mathbf{x}_1\in\mathbb{R}^n$, $\mathbf{x}_2\in\mathbb{C}^n$ and $\mathbf{x}_2+\mathbf{x}_1$ where $\mathbf{x}_2$ and $\mathbf{x}_1$ are linearly independent. Then for Eigen values $\lambda_1, \lambda_2, \lambda_3\in\mathbb{C}$,

$\mathbf{Ax}_1=\lambda_1\mathbf{x}_1$

$\mathbf{Ax}_2=\lambda_2\mathbf{x}_2$

$\mathbf{A}\left(\mathbf{x}_1+\mathbf{x}_2\right)=\lambda_3\left(\mathbf{x}_1+\mathbf{x}_2\right)$

$\Rightarrow\left(\lambda_3-\lambda_1\right)\mathbf{x}_1=\left(\lambda_2-\lambda_3\right)\mathbf{x}_2$

From the linear independence condition, $\lambda_3=\lambda_1=\lambda_2$ and hence any vector $\mathbf{x}$ in the subspace spanned by $\mathbf{x}_1$ and $\mathbf{x}_2$ is an Eigen vector. See geometric multiplicity and Eigen space.

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