关于“n个independent list of vector的span中任意n+1个vector构成的list必为dependent”的证明动机及简化证明问询

Hey there, I totally get where you're coming from—Apostol's proofs can feel like they're pulling tricks out of thin air sometimes, even when you follow each step individually. Let's break this down first by unpacking the motivation behind those case splits in Apostol's proof, then I'll walk you through a simpler proof that sticks to basic linear algebra concepts you mentioned (linear spaces, subspaces, span, linear independence).

先聊聊Apostol证明里的动机（Case 1及后续） #

Apostol’s approach is usually built on induction, right? Let’s assume he’s using induction on the dimension ( n ) here.

• Case 1 is probably when the first vector of the n+1 list is in the span of the original n independent vectors—the motivation here is to "reduce" the problem: if that first vector can be written as a combination of the original list, we can substitute it out, then use the induction hypothesis on the remaining n vectors (since we’ve now linked the n+1 list to the original n independent set).
• The follow-up cases (like if the first vector isn’t in the span) would be about leveraging the definition of linear independence: adding a vector not in the span of an independent list gives a larger independent list—but wait, we only have n dimensions, so that’s a contradiction. That’s the core intuition, even if Apostol’s wording makes it feel circuitous.

The reason he splits into cases is to cover all possibilities systematically, which is standard in formal proofs, but it can obscure the "big picture" intuition that we’re just using the fact that the span of n independent vectors has dimension n—so you can’t have more than n independent vectors in it.

简化版证明（仅用线性空间、子空间、张成、线性无关的定义） #

Let’s start with the formal statement to be clear:

Let ( V ) be a linear space, let ( S = {v_1, v_2, ..., v_n} ) be a linearly independent list of vectors in ( V ), and let ( W = \text{span}(S) ). Any list ( T = {w_1, w_2, ..., w_{n+1}} ) of vectors in ( W ) is linearly dependent.

Here’s the step-by-step proof:

1. Express each ( w_i ) as a linear combination of ( S )
   Since every ( w_i \in W = \text{span}(S) ), we can write:

   w_1 = a_{11}v_1 + a_{12}v_2 + ... + a_{1n}v_n
   w_2 = a_{21}v_1 + a_{22}v_2 + ... + a_{2n}v_n
   ...
   w_{n+1} = a_{(n+1)1}v_1 + a_{(n+1)2}v_2 + ... + a_{(n+1)n}v_n
   

   Where each ( a_{ij} ) is a scalar from the field of ( V ).

2. Set up a linear dependence equation for ( T )
   We want to find scalars ( c_1, c_2, ..., c_{n+1} ), not all zero, such that:

   c_1w_1 + c_2w_2 + ... + c_{n+1}w_{n+1} = 0
   

3. Substitute the combinations from step 1 into the equation
   Replace each ( w_i ) with its expression in terms of ( S ):

   c_1(a_{11}v_1 + ... + a_{1n}v_n) + c_2(a_{21}v_1 + ... + a_{2n}v_n) + ... + c_{n+1}(a_{(n+1)1}v_1 + ... + a_{(n+1)n}v_n) = 0
   

   Rearrange terms by grouping coefficients of each ( v_j ):

   (c_1a_{11} + c_2a_{21} + ... + c_{n+1}a_{(n+1)1})v_1 + ... + (c_1a_{1n} + c_2a_{2n} + ... + c_{n+1}a_{(n+1)n})v_n = 0
   

4. Use linear independence of ( S )
   Since ( {v_1, ..., v_n} ) is linearly independent, the coefficient of each ( v_j ) must be zero. This gives us a system of n linear equations with n+1 variables (( c_1, ..., c_{n+1} )):

   a_{11}c_1 + a_{21}c_2 + ... + a_{(n+1)1}c_{n+1} = 0
   a_{12}c_1 + a_{22}c_2 + ... + a_{(n+1)2}c_{n+1} = 0
   ...
   a_{1n}c_1 + a_{2n}c_2 + ... + a_{(n+1)n}c_{n+1} = 0
   

5. Conclude the system has a non-trivial solution
   A homogeneous system of linear equations with fewer equations than variables always has at least one non-trivial solution (i.e., not all ( c_i = 0 )). That’s a key result that ties directly to our core concepts: you can’t have more "free" variables than constraints without getting a non-zero solution, which aligns with the idea that the span of n independent vectors can’t support more than n independent vectors.

6. Final step: non-trivial solution implies linear dependence
   The non-trivial scalars ( c_1, ..., c_{n+1} ) we found satisfy the equation ( c_1w_1 + ... + c_{n+1}w_{n+1} = 0 ), so the list ( T ) is linearly dependent.

That’s it! This proof sticks strictly to the concepts you mentioned—no fancy tricks, just following the definitions and using the intuitive result about underdetermined systems.

备注：内容来源于stack exchange，提问作者Champayond