= Solution
Two nonzero vectors are <linearly independent> when $s\mathbf x+t\mathbf y=0$ forces $s=t=0$. Their <linear span> then has dimension two. If they are <linearly dependent>, one is a nonzero scalar multiple of the other, and their span has dimension one. Thus the two requested dimensions are $\boxed{2\text{ and }1}$.
The Euclidean <scalar product> and its <norm> are $\mathbf x\cdot\mathbf y=\sum_{j=1}^n x_jy_j$ and $\|\mathbf x\|=\sqrt{\mathbf x\cdot\mathbf x}$. To prove the <Cauchy-Schwarz inequality>, first handle $\mathbf y=0$ trivially. Otherwise the squared <norm>
$$
0\leq\left\|\mathbf x-\frac{\mathbf x\cdot\mathbf y}{\|\mathbf y\|^2}\mathbf y\right\|^2
=\|\mathbf x\|^2-\frac{(\mathbf x\cdot\mathbf y)^2}{\|\mathbf y\|^2}
$$
gives
$$
\boxed{|\mathbf x\cdot\mathbf y|\leq\|\mathbf x\|\|\mathbf y\|.}
$$
Equality holds exactly when the residual vector is zero, meaning the vectors are <linearly dependent>; this includes the zero-vector cases. Expanding $\|\mathbf x+\mathbf y\|^2$ and applying <Cauchy-Schwarz> then gives
$$
\|\mathbf x+\mathbf y\|^2\leq\|\mathbf x\|^2+2\|\mathbf x\|\|\mathbf y\|+\|\mathbf y\|^2,
\qquad\boxed{\|\mathbf x+\mathbf y\|\leq\|\mathbf x\|+\|\mathbf y\|}.
$$
Taking nonnegative square roots proves the <triangle inequality>.
For the unit-vector optimization, let $\mathbf w=\mathbf x+\mathbf y$. Independence ensures $\mathbf w\ne0$. The variable part of $S$ is $\mathbf z\cdot\mathbf w$, which <Cauchy-Schwarz> bounds below by $-\|\mathbf w\|$. Equality occurs only for the antiparallel unit vector. Thus
$$
\boxed{\mathbf z_*=-\frac{\mathbf x+\mathbf y}{\|\mathbf x+\mathbf y\|},\qquad
\lambda=-\frac1{\|\mathbf x+\mathbf y\|},\qquad
S_{\min}=\mathbf x\cdot\mathbf y-\sqrt{2+2\mathbf x\cdot\mathbf y}.}
$$
This is <minimizing a linear functional on a sphere>. The two consequences follow below.
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