# Convex combinations of $m$ vectors is a minimal affine subset and spans $m-1$ dimensions

This exercise is from Linear Algebra Done Right by Sheldon Axler (Chapter 3.E, exercise number 11). I tried to figure out a solution and I would like to have some hints.

Suppose $$v_1,\dots v_m \in V$$. Let $$A = \{ \lambda_1v_1 + \dots + \lambda_mv_m : \lambda_1,\dots,\lambda_m \in F \text{ and } \lambda_1 + \dots + \lambda_m = 1\}$$

(a) Prove that $$A$$ is an affine subset of $$V$$.

(b) Prove that every affine subset of $$V$$ that contains $$v_1,\dots,v_m$$ also contains $$A$$.

(c) Prove that $$A = v + U$$ for some $$v \in V$$ and some subspace $$U$$ of $$V$$ with $$\dim U \le m-1$$.

My attempts

Let's start by saying that a non empty subset A of a space vector V over the field F is an affine subset of V if and only if λv + (1-λ)w ∈ A for all v, w ∈ A and λ ∈ F. We can prove (a): if $$x_1$$, $$x_2$$ ∈ A, α($$x_1$$) + (1-α)($$x_2$$) ∈ A, for α ∈ F. This implies that α($$λ_1v_1$$+…+$$λ_mv_m$$) + (1-α)($$β_1v_1$$+…+$$β_mv_m$$) --> α$$λ_1v_1$$+…+α$$λ_mv_m$$ + $$β_1v_1$$+…+$$β_mv_m$$...-α$$β_1v_1$$…-α$$β_mv_m$$; let's consider that α$$λ_1$$+…+α$$λ_m$$ = α, α$$β_1$$+…+α$$β_m$$ = α; so we have that α($$x_1$$) + (1-α)($$x_2$$) = $$β_1v_1$$+…+$$β_mv_m$$ ∈ A. So A is an affine subset. ($$β_1v_1$$+…+$$β_mv_m$$ = 1).

Point (b): let Z an affine subset of V and $$v_1$$,..,$$v_m$$ ∈ Z. For the implication above, γ$$v_1$$ + (1-γ)$$v_2$$ ∈ Z. For the same reason, δγ$$v_1$$ + δ(1-γ)$$v_2$$ + (1-δ)$$v_3$$ ∈ Z. We can repeat for all $$v_1$$,...,$$v_m$$ until it’ll result that some linear combinations of $$v_1$$,…,$$v_m$$ belong to Z. A is one of that by point (a).

Regarding point (c): we now know that A is an affine subspace. Let’s call U the subset U of V such that A = U + v, for some v ∈ V. We can say that $$λ_1v_1$$ +…+$$λ_mv_m$$ – v ∈ U. With $$λ_1v_1$$+…+$$λ_mv_m$$ ∉ U, we can deduce that it’s possible to reduce $$v_1$$,…$$v_m$$ to a spanning list of U by subtracting one more vectors. So the spanning list will be compound by not more than (m-1) vectors. We can reduce the spanning list to a basis; in the end, dim(U) ≤ (m-1).

• Please use mathjax. It provides intuitive constructs for the more common notations, such as $v_1$ for subscripts $v_1$ and likewise $x^2$ for superscripts $x^2$. If more than a single letter is raised or lowered, use grouping $a^{12}$ to get $a^{12}$. Oct 13, 2023 at 21:21
• @LutzLehmann thank you for the advice Oct 13, 2023 at 22:01
• I’ve edited the question for you. Please edit your attempts similarly for future readers’ sake. You can check Detexify for symbols’ names in LaTeX. Oct 14, 2023 at 15:24

An affine subset is defined as a subset of $$V$$ of the form $$v+U$$ for some $$v \in V$$ and some subspace $$U$$ of $$V$$.

$$(a)$$ For part $$(a)$$, we can prove that $$A=\left \{\sum \lambda_iv_i: \sum \lambda_i = 1 \right\}$$ is an affine subset by finding some $$v \in V$$ and some subspace $$U$$ of $$V$$ such that $$A=v+U.$$

Naturally $$v$$ and $$U$$ are related to $$v_i$$'s.

If we choose $$v=v_1$$, then for any $$u \in U$$,

$$v_1+u=\lambda_1v_1+\lambda_2v_2+\cdots+\lambda_mv_m \; \mathrm{where} \; \sum_{i=1}^{m} \lambda_i = 1$$

$$\implies u = (\lambda_1-1)v_1+\lambda_2v_2+\cdots+\lambda_mv_m \; \mathrm{where} \; \sum_{i=1}^{m} \lambda_i = 1$$

Notice that $$(\lambda_1-1)+\lambda_2+\cdots+\lambda_m=\sum_{i=1}^{m} \lambda_i -1 = 0$$ $$\therefore$$ if we set $$\alpha_1=\lambda_1-1, \alpha_i=\lambda_i \; \mathrm{for} \; i=2, \cdots, m,$$ then $$u=\sum_{i=1}^m \alpha_i v_i \;\mathrm{where} \; \sum_{i=1}^{m} \alpha_i = 0$$

In conclusioin, for part (a), we may try to prove that $$U=\left \{\sum_{i=1}^m \alpha_i v_i : \sum_{i=1}^{m} \alpha_i = 0 \right\}$$ is a subspace of $$V$$ and $$A=v_1+U.$$

$$(b)$$ Suppose that $$x+W$$ is an affine subset containing $$v_1, v_2, ..., v_m$$, then for $$i=1, 2, \cdots, m,$$ $$v_i=x+w_i \; \mathrm{for \; some} \; w_i \in W.$$

What we need to show is that if $$\; \sum_{i=1}^{m} \lambda_i = 1$$, then $$\sum_{i=1}^{m} \lambda_iv_i \in x+W.$$

$$(c)$$ In $$(a)$$, we have already found an $$U$$ such that $$A=v_1+U$$ .

Therefore for this part, we only need to prove that $$\dim U \leq m-1.$$

Since $$U \subset \mathrm{span}\left(v_1, v_2, \cdots, v_m \right)$$ and $$\dim \mathrm{span}\left(v_1, v_2, \cdots, v_m \right) \leq m,$$ we are done if we can prove that $$U \neq \mathrm{span}\left(v_1, v_2, \cdots, v_m \right).$$