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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_1$$v_1$+…+$λ_m$$v_m$) + (1-α)($β_1$$v_1$+…+$β_m$$v_m$) --> α$λ_1$$v_1$+…+α$λ_m$$v_m$ + $β_1$$v_1$+…+$β_m$$v_m$...-α$β_1$$v_1$…-α$β_m$$v_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_1$$v_1$+…+$β_m$$v_m$ ∈ A. So A is an affine subset. ($β_1$$v_1$+…+$β_m$$v_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_1$$v_1$ +…+$λ_m$$v_m$ – v ∈ U. With $λ_1$$v_1$+…+$λ_m$$v_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).

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    $\begingroup$ 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 $a^{12}$. $\endgroup$ Oct 13, 2023 at 21:21
  • $\begingroup$ @LutzLehmann thank you for the advice $\endgroup$
    – datfq
    Oct 13, 2023 at 22:01
  • $\begingroup$ 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. $\endgroup$ Oct 14, 2023 at 15:24

1 Answer 1

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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 $$U \neq \mathrm{span}\left(v_1, v_2, \cdots, v_m \right).$$

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