4
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I'm working on a problem from my textbook and found that $\left(\frac{1}{2}, \frac{1}{2}, 1\right)$ is an eigenvector for a particular eigenvalue of $4$.

The textbook solution says that the answer is $(1, 1, 2)$ which is just $2 \times \left(\frac{1}{2}, \frac{1}{2}, 1\right)$

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6
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Yes it is. Here is a short proof

Take $Ax = \lambda x$ and multiply by scalar $k$ we get $kAx = k \lambda x$

but $kAx$ = $A(kx)$ and $ k \lambda x$ = $\lambda (kx)$ so we see by definition that $kx$ is an eigenvector

$$A(kx) = \lambda (kx)$$

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  • $\begingroup$ Thanks so much! I'm pretty sure that's the definition of an eigenvector from my textbook too. lol $\endgroup$ – Clark Bell Dec 3 '15 at 4:50

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