Math Problem Statement

Suppose 𝑣1 , … , π‘£π‘š is a list of vectors in 𝑉. For π‘˜ ∈ {1, … , π‘š}, let π‘€π‘˜ = 𝑣1 + β‹― + π‘£π‘˜ . Show that span(𝑣1 , … , π‘£π‘š) = span(𝑀1 , … , π‘€π‘š).

Solution

We are tasked with proving that span(v1,v2,…,vm)=span(w1,w2,…,wm)\text{span}(v_1, v_2, \dots, v_m) = \text{span}(w_1, w_2, \dots, w_m), where wk=v1+v2+β‹―+vkw_k = v_1 + v_2 + \cdots + v_k for each k∈{1,2,…,m}k \in \{1, 2, \dots, m\}.

Step 1: Recall Definitions

  • span(v1,v2,…,vm)\text{span}(v_1, v_2, \dots, v_m) refers to the set of all possible linear combinations of the vectors v1,v2,…,vmv_1, v_2, \dots, v_m.

    In other words: span(v1,v2,…,vm)={Ξ±1v1+Ξ±2v2+β‹―+Ξ±mvm∣α1,Ξ±2,…,Ξ±m∈R}.\text{span}(v_1, v_2, \dots, v_m) = \left\{ \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_m v_m \mid \alpha_1, \alpha_2, \dots, \alpha_m \in \mathbb{R} \right\}.

  • Similarly, span(w1,w2,…,wm)\text{span}(w_1, w_2, \dots, w_m) refers to the set of all possible linear combinations of the vectors w1,w2,…,wmw_1, w_2, \dots, w_m, where each wk=v1+v2+β‹―+vkw_k = v_1 + v_2 + \cdots + v_k.

    So, span(w1,w2,…,wm)={Ξ²1w1+Ξ²2w2+β‹―+Ξ²mwm∣β1,Ξ²2,…,Ξ²m∈R}.\text{span}(w_1, w_2, \dots, w_m) = \left\{ \beta_1 w_1 + \beta_2 w_2 + \cdots + \beta_m w_m \mid \beta_1, \beta_2, \dots, \beta_m \in \mathbb{R} \right\}.

Step 2: Show span(v1,…,vm)βŠ†span(w1,…,wm)\text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m)

Each vector vkv_k can be expressed as a linear combination of w1,…,wkw_1, \dots, w_k because:

  • w1=v1w_1 = v_1,
  • w2=v1+v2w_2 = v_1 + v_2, so v2=w2βˆ’w1v_2 = w_2 - w_1,
  • w3=v1+v2+v3w_3 = v_1 + v_2 + v_3, so v3=w3βˆ’w2v_3 = w_3 - w_2,
  • In general, for any kk, we can express: vk=wkβˆ’wkβˆ’1.v_k = w_k - w_{k-1}.

This shows that each vector vkv_k can be written as a linear combination of the vectors w1,w2,…,wkw_1, w_2, \dots, w_k, implying that any linear combination of v1,v2,…,vmv_1, v_2, \dots, v_m can also be written as a linear combination of w1,w2,…,wmw_1, w_2, \dots, w_m. Thus: span(v1,v2,…,vm)βŠ†span(w1,w2,…,wm).\text{span}(v_1, v_2, \dots, v_m) \subseteq \text{span}(w_1, w_2, \dots, w_m).

Step 3: Show span(w1,…,wm)βŠ†span(v1,…,vm)\text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m)

Next, we show that each wkw_k is a linear combination of v1,v2,…,vkv_1, v_2, \dots, v_k. Specifically, by definition: wk=v1+v2+β‹―+vk.w_k = v_1 + v_2 + \cdots + v_k. This is already a linear combination of the vectors v1,v2,…,vkv_1, v_2, \dots, v_k, meaning that any linear combination of w1,w2,…,wmw_1, w_2, \dots, w_m is also a linear combination of v1,v2,…,vmv_1, v_2, \dots, v_m. Thus: span(w1,w2,…,wm)βŠ†span(v1,v2,…,vm).\text{span}(w_1, w_2, \dots, w_m) \subseteq \text{span}(v_1, v_2, \dots, v_m).

Step 4: Conclude the Equality

Since we have shown both: span(v1,v2,…,vm)βŠ†span(w1,w2,…,wm)\text{span}(v_1, v_2, \dots, v_m) \subseteq \text{span}(w_1, w_2, \dots, w_m) and span(w1,w2,…,wm)βŠ†span(v1,v2,…,vm),\text{span}(w_1, w_2, \dots, w_m) \subseteq \text{span}(v_1, v_2, \dots, v_m), we can conclude that: span(v1,v2,…,vm)=span(w1,w2,…,wm).\text{span}(v_1, v_2, \dots, v_m) = \text{span}(w_1, w_2, \dots, w_m).

Final Result:

Thus, we have shown that span(v1,…,vm)=span(w1,…,wm)\text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m).

Would you like further clarifications, or have any questions about this proof?

Here are 5 related questions you might find interesting:

  1. What is the geometric interpretation of spanning sets in R2\mathbb{R}^2 and R3\mathbb{R}^3?
  2. How does the dimensionality of a vector space relate to its spanning set?
  3. Can two different sets of vectors span the same subspace in a vector space?
  4. What role do linear independence and dependence play in spanning sets?
  5. How does the concept of a basis relate to the span of a set of vectors?

Tip: To check if a set of vectors spans a space, determine if their linear combinations can generate any vector in the space.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Spanning Sets

Formulas

span(v1, v2, ..., vm) = {Ξ±1v1 + Ξ±2v2 + ... + Ξ±mvm | Ξ±1, Ξ±2, ..., Ξ±m ∈ ℝ}
wk = v1 + v2 + ... + vk

Theorems

Span of a Set of Vectors
Linear Combinations

Suitable Grade Level

Undergraduate Level (College)