= Solution
If $S$ spans and has $n$ elements but is <linearly dependent>, a nontrivial <linear combination> equal to zero lets us remove one member without changing the <span>. This produces $n-1$ spanning vectors, whereas the <standard basis> contains $n$ <linearly independent> vectors, contradicting the <Steinitz exchange lemma>. Therefore $\boxed{S\text{ is linearly independent}}$. Together the three deductions establish that any two of the three properties imply the third. For $n=0$, the only independent or zero-element family is the empty <basis>, and the same conclusions hold.
Relative to the given <basis> $e_1,e_2$, the proposed vectors are the columns of
$$
M=\begin{pmatrix}\lambda&1\\1&\lambda\end{pmatrix},\qquad \det M=\lambda^2-1.
$$
The <determinant> is nonzero exactly when the columns form a <basis>. Thus $\boxed{\lambda\ne\pm1}$. At $\lambda=1$ the vectors coincide, and at $\lambda=-1$ they are negatives of one another.
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