Exercise 3.E.11

Answers

(a) By problem 3.E.8, $A$ is an affine subset if and only if $ty + (1 - t) x \in A$ for $x, y \in A$ and $t \in F$
Define $x, y \in A$ :

x = a_{1} v_{1} + \dots + a_{m} v_{m}

y = b_{1} v_{1} + \dots + b_{m} v_{m}

Where

a_{1} + \dots + a_{m} = 1

and,

b_{1} + \dots + b_{m} = 1

Now, let

z = (1 - t) x + ty

Clearly, $z$ is in the form in 3.E.8, so we just need to prove that $z \in A$
So,

\begin{matrix} \begin{aligned} (1 - t) x + ty & = (1 - t) (a_{1} v_{1} + \dots + a_{m} v_{m}) + t (b_{1} v_{1} + \dots + b_{m} v_{m}) \\ = (1 - t) a_{1} v_{1} + t (b_{1} v_{1}) + \dots + (1 - t) a_{m} v_{m} + t b_{m} v_{m} \\ = ((1 - t) a_{1} + t b_{1}) v_{1} + \dots + ((1 - t) a_{m} + t b_{m}) v_{m} \end{aligned} \end{matrix}

(1)

Now define

λ_{j} = (1 - t) a_{j} + t b_{j}

Since, $λ_{1} + \dots + λ_{m} = 1$ , $z \in A$ , and thus, $A$ is an affine subset.

(b) Define $U \subset V$ and $b \in V$ Next, define $B$ an affine subset of $V$ , where

B = b + U

such that

v_{1}, . . ., v_{m} \in b + U

Define $u_{1}, . . ., u_{m} \in U$ such that

v_{j} = b + u_{j}

For $j \in {1, . . ., m}$ Therefore,

\begin{matrix} \begin{aligned} λ_{1} b + λ_{1} + u_{1} + \dots + λ_{m} b + λ_{m} u_{m} & = b + λ_{1} u_{1} + \dots + λ \\ \in b + U \end{aligned} \end{matrix}

(2)

a = λ_{1} v_{1} + + λ_{m} v_{m}

Where $λ_{1} + + λ_{m} = 1$ Now, define $u \in V$ , s.t.

u = a - v_{1} = λ_{1} v_{1} + \dots λ_{m} v_{m} - v_{1}

(3)

Since $v_{1} = λ_{1} v_{1} + \dots λ_{m} + v_{1}$ ,

λ_{1} v_{1} + \dots λ_{m} v_{m} - v_{1} = λ_{2} (v_{2} - v_{1}) + \dots + λ_{m} (v_{m} - v_{1})

Now, define $U = span (v_{2} - v_{1}, . . ., v_{m} - v_{1})$
Obviously, $A \subset v_{1} + U$ by (3)
Therefore, $dim A \leq dim U$
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Arden_Shi

2025-06-08 14:53