Solution
= Solution
Because $\mathbf b\cdot\mathbf x$ is a <scalar>, applying the <linear operator> again gives
$$
K^2\mathbf x=K\bigl[\mathbf a(\mathbf b\cdot\mathbf x)\bigr]=\mathbf a\bigl[\mathbf b\cdot\mathbf a(\mathbf b\cdot\mathbf x)\bigr]=(\mathbf a\cdot\mathbf b)K\mathbf x.
$$
This is the first nontrivial case of the <powers of a rank-one operator>; its <matrix> is the <outer product> $\mathbf a\mathbf b^T$.