Math Problem Statement
Suppose π£1 , β¦ , π£π is a list of vectors in π. For π β {1, β¦ , π}, let π€π = π£1 + β― + π£π . Show that span(π£1 , β¦ , π£π) = span(π€1 , β¦ , π€π).
Solution
We are tasked with proving that , where for each .
Step 1: Recall Definitions
-
refers to the set of all possible linear combinations of the vectors .
In other words:
-
Similarly, refers to the set of all possible linear combinations of the vectors , where each .
So,
Step 2: Show
Each vector can be expressed as a linear combination of because:
- ,
- , so ,
- , so ,
- In general, for any , we can express:
This shows that each vector can be written as a linear combination of the vectors , implying that any linear combination of can also be written as a linear combination of . Thus:
Step 3: Show
Next, we show that each is a linear combination of . Specifically, by definition: This is already a linear combination of the vectors , meaning that any linear combination of is also a linear combination of . Thus:
Step 4: Conclude the Equality
Since we have shown both: and we can conclude that:
Final Result:
Thus, we have shown that .
Would you like further clarifications, or have any questions about this proof?
Here are 5 related questions you might find interesting:
- What is the geometric interpretation of spanning sets in and ?
- How does the dimensionality of a vector space relate to its spanning set?
- Can two different sets of vectors span the same subspace in a vector space?
- What role do linear independence and dependence play in spanning sets?
- 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.
Ask a new question for Free
By Image
Drop file here or Click Here to upload
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 β β}
wk = v1 + v2 + ... + vk
Theorems
Span of a Set of Vectors
Linear Combinations
Suitable Grade Level
Undergraduate Level (College)
Related Recommendation
Understanding Span Equality in Vector Spaces
Proof that span(v1, ..., vm) = span(w1, ..., wm)
Linear Algebra Proof: Spanning Sets in R^n for u, v, w
Proof of Span Equality in Vector Spaces: span {u, v, w} = span {u, u + v, u + v + w}
Proof via Case Distinction: Dimension of Span of Vectors with Common Component w