Vectors Generating R^3 Linear Algebra Problem

Published on September 15, 2025

Linear-Algebra

Determine if a set of vectors generates R^3 by checking for linear independence or solving a system of linear equations.

Question

Answer

Here are the solutions to the problems:
Problem 2:
In , let . Is a subspace of ?
To determine if is a subspace of , we need to check three conditions:
  1. The zero vector must be in .
  2. must be closed under vector addition.
  3. must be closed under scalar multiplication.
Let be the vector space of polynomials of degree at most 4 with real coefficients.
  1. Zero vector: The zero polynomial is for all .
    For the zero polynomial, .
    Thus, the zero polynomial is in .
  2. Closure under addition: Let and .
    By definition of , and .
    Consider their sum .
    We need to check if .
    .
    So, . is closed under addition.
  3. Closure under scalar multiplication: Let and be a scalar.
    By definition of , .
    Consider the scalar product .
    We need to check if .
    .
    So, . is closed under scalar multiplication.
Since all three conditions are met, is a subspace of .
Answer for Problem 2: Yes, is a subspace of .

Problem 3:
Let . Is a vector subspace of ?
To determine if is a subspace of , we need to check the three conditions:
  1. The zero vector must be in .
  2. must be closed under vector addition.
  3. must be closed under scalar multiplication.
  4. Zero vector: The zero vector in is .
    For , we have .
    Thus, the zero vector is in .
  5. Closure under addition: Let’s test this condition.
    Consider two vectors in , for example, and .
    For : , so .
    For : , so .
    Now consider their sum .
    For : .
    Since , the vector is not in .
    Therefore, is not closed under addition.
Since is not closed under addition, it cannot be a subspace of . (We don’t even need to check scalar multiplication, but for completeness, let’s check it).
  1. Closure under scalar multiplication: Let’s test this condition.
    Consider a vector in , for example, .
    For : , so .
    Let be a scalar.
    Consider the scalar product .
    For : .
    Since , the vector is not in .
    Therefore, is not closed under scalar multiplication.
Answer for Problem 3: No, is not a vector subspace of .

Problem 4:
Determine if the set generates .
A set of vectors generates if any vector can be written as a linear combination of the vectors in the set . That is, we need to find scalars such that:
This gives us a system of linear equations:


Now we can solve this system:
From the third equation, .
Substitute into the first equation: .
Substitute into the second equation: .
Since we can always find unique values for for any given , the set generates .
Alternatively, we can check if the vectors are linearly independent. If they are, and there are 3 vectors in , then they form a basis and thus generate .
We can form a matrix with these vectors as rows or columns and calculate its determinant.
Let the matrix be:
The determinant of is:


Since the determinant is non-zero ( ), the vectors are linearly independent.
Since there are 3 linearly independent vectors in (which has dimension 3), they form a basis for and therefore generate .
Answer for Problem 4: Yes, the set generates .