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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 ofThe determinant is nonzero exactly when the columns form a basis. Thus $\boxed{\lambda \ne \pm 1}$. At $\lambda =1$ the vectors coincide, and at $\lambda =-1$ they are negatives of one another.

If $S$ is linearly independent and has $n$ elements, apply the Steinitz exchange lemma with $S$ as the independent family and the standard basis as the spanning family. All $n$ members of the standard basis are replaced, leaving $S$ itself spanning. Thus $\boxed{\mathrm{span}S={\mathbb{R}}^n}$, so $S$ is a basis.

The Steinitz exchange lemma says that if $w_1,\dots ,w_m$ span a vector space and $v_1,\dots ,v_r$ are linearly independent, then $r\le m$, and after reordering the $w_j$, the family $v_1,\dots ,v_r,w_{r+1},\dots ,w_m$ still spans. Here is a proof. Suppose the first $k-1$ replacements have been made. Express $v_k$ in that spanning family. Some coefficient of a remaining $w_j$ must be nonzero, since otherwise $v_k$ belongs to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Solve for that $w_j$ and replace it by $v_k$; this preserves spanning. If $r>m$, the first $m$ replacements leave a spanning family $v_1,\dots ,v_m$, contradicting linear independence of the first $m+1$ vectors. This proves both assertions.

Now suppose $S$ is linearly independent and spans ${\mathbb{R}}^n$. Express each member of the standard basis using finitely many members of $S$, and let $S_0$ be the finite union of those members. Then $S_0$ spans. No element of $S\setminus S_0$ can exist, since it would be a linear combination of $S_0$ and violate linear independence. Thus $S$ is finite. Applying the Steinitz exchange lemma in both directions to $S$ and the standard basis gives $|S|\le n$ and $n\le |S|$. Hence $\boxed{|S|=n}$.

Suppose $y$ in the capped simplex has a coordinate strictly between zero and one. Since ${\sum }_i{y}_i=k$ is an integer, it cannot have exactly one such coordinate: the other coordinates would contribute an integer sum. Hence there are distinct $i,j$ with $0<y_i,y_j<1$.

Choose $0<\epsilon <\min \{y_i,1-y_i,y_j,1-y_j\}$ and set $y^{\pm }=y\pm \epsilon (e_i-e_j)$, where $e_i,e_j$ are standard basis vectors. Both perturbed vectors satisfy the coordinate bounds and have the same coordinate sum. They are distinct and $y=(y^{+}+y^{-})/2$, so $y$ is not an extreme point.

Conversely, every zero-one vector in this convex polytope is an extreme point. If it were a nontrivial convex combination of two feasible vectors, each coordinate equal to zero would force both corresponding coordinates to be zero, and each coordinate equal to one would force both to be one. Thus the two vectors would equal the original one. The extreme points are therefore exactly the indicators of $k$-element subsets.

For algebraic stability, the weights must be nonnegative and the real symmetric matrix $M=BA+A^{T}B-b{b}^T$, where $B=\mathrm{diag}(b)$, must be a positive semidefinite matrix.

The weights are positive, but $a_{11}=0$ and $b_1=1/6$, soA positive semidefinite matrix cannot have a negative diagonal entry, as its quadratic form at the first standard basis vector would be negative. Thus this Lobatto IIIA method is not algebraically stable, despite its A-stability.