Is the scalar multiple of an eigenvector also an eigenvector for a particular eigenvalue?

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 $2 \times \left(\frac{1}{2}, \frac{1}{2}, 1\right)$

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