Question

We often write vectors in \(\mathbb{R}^{n}\) as rows. Is it possible that \\{(1,2,0),(2,0,3)\\} can span the subspace \(U=\\{(r, s, 0) \mid r\) and \(s\) in \(\mathbb{R}\\} ?\) Defend your answer.

Short answer

No, the set cannot span \( U \) because \( (2,0,3) \) forces restrictive conditions that cannot express all \((r, s, 0)\).

Step-by-step solution

Understand the Problem

We need to determine whether the set \( \{(1,2,0), (2,0,3)\} \) can span the subspace \( U = \{(r, s, 0) \mid r \text{ and } s \in \mathbb{R}\} \). This subspace includes all vectors in \( \mathbb{R}^3 \) with the form \((r, s, 0)\).

Identify the Form of Vectors in Subspace U

The subspace \( U \) is defined as \( \{(r, s, 0) \mid r \text{ and } s \in \mathbb{R}\} \). Any vector in this subspace has a zero as its third component.

Analyze the Given Set of Vectors

The set contains the vectors \((1, 2, 0)\) and \((2, 0, 3)\). Notice that \((1, 2, 0)\) belongs to the subspace \( U \) since its third component is zero, but \((2, 0, 3)\) does not belong to \( U \) as its third component is 3.

Determine if the Set Can Span the Subspace

To span \( U \), any vector from \( U \) such as \((r, s, 0)\) should be expressible as a linear combination of \((1,2,0)\) and \((2,0,3)\). So, we must have coefficients \( a \) and \( b \) such that \( a(1,2,0) + b(2,0,3) = (r,s,0) \).

Set Up the Linear Combination for Spanning

From above, we get the system:1. \( a + 2b = r \)2. \( 2a = s \)3. \( 3b = 0 \)

Solve the System of Equations

Equation 3, \( 3b = 0 \), implies \( b = 0 \). Substitute \( b = 0 \) into the other equations:- Equation 1 becomes \( a = r \).- Equation 2 becomes \( 2a = s \) or \( a = \frac{s}{2}\).For these to both hold for any \( r \) and \( s \), they contradict each other unless \( s = 2a \) also equals \( 2r \), which is not necessarily true for all \( r \) and \( s \).

Conclusion Based on the Analysis

Since \((2, 0, 3)\) contributed a condition \( b = 0 \) that limits the expression of \( (r, s, 0) \) using \((1, 2, 0)\) and itself, it cannot cover all combinations in \( U \). Hence, the set cannot span \( U \).

Key concepts

Linear Combinations

In the world of vector spaces, a linear combination involves combining a set of vectors using scalar multiplication and addition. Imagine you have two vectors, say \((1, 2, 0)\) and \((2, 0, 3)\). You can create a linear combination of these vectors by multiplying each vector by a scalar (let's call them \(a\) and \(b\) respectively) and then adding the results. This is represented as \(a(1, 2, 0) + b(2, 0, 3)\). This process gives us a new vector that is formed from the initial set. Linear combinations are essential to understand because they tell us what new vectors we can form from a given set of vectors. If we can express every vector in a particular space as a linear combination of a set of vectors, these vectors span the space. But as we will see, not all sets of vectors can span every possible space of interest.

Subspace

A subspace is like a smaller home within a larger neighborhood, where this neighborhood is a vector space. Formally, a subspace is any set of vectors that is closed under vector addition and scalar multiplication. This means that if you add any two vectors in the subspace, you still get another vector in the subspace. Similarly, if you multiply any vector in the subspace by a scalar, the result is still inside the subspace.In our context, the subspace \( U = \{(r, s, 0) \mid r \text{ and } s \in \mathbb{R}\} \) represents all vectors that have the form \((r, s, 0)\) in three-dimensional space \(\mathbb{R}^3\). This will include infinite combinations of \(r\) and \(s\), but crucially, the third component is always zero. The condition of having a zero in the third component defines the constraint for any vector to belong to this subspace.

System of Equations

When we try to check if a set of vectors can span a subspace, one approach is to set up what's called a system of equations. This system helps us see if every vector in the subspace can be written as a linear combination of the given vectors. In our exercise, the system derived from the linear combination is:1. \( a + 2b = r \) 2. \( 2a = s \) 3. \( 3b = 0 \)Here, each equation represents a condition that needs to be met for the linear combination to match a vector in the subspace for any values of \( r \) and \( s \). Solving this system reveals if such vectors can or cannot express every possible vector \((r, s, 0)\) in our subspace. The third equation is particularly crucial here since it dictates \( b = 0 \), leading to constraints that can't necessarily cover the entire subspace.

Spanning Set

A spanning set is a collection of vectors from which any vector in a given space can be formed using a linear combination of the set. If you have a set of vectors that spans a space, it essentially means these vectors can reach every corner of that space through various combinations.In our example with vectors \((1, 2, 0)\) and \((2, 0, 3)\), we need to check if they can span the subspace \( U = \{(r, s, 0) \mid r \text{ and } s \in \mathbb{R}\} \). Upon setting up and solving the system of equations, we find a restriction: with \( b \) being zero, the contribution of the vector \((2, 0, 3)\) becomes irrelevant in reaching vectors like \((r, s, 0)\) with non-zero \(r\) or \(s\). Therefore, this set cannot generate every vector in our targeted subspace, failing to be a spanning set for \( U \). Understanding the properties and limitations of spanning sets is critical as it guides the capability of vectors to fully characterize a space.