Question

We often write vectors in \(\mathbb{R}^{n}\) as rows. Suppose that \(U=\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) where each \(\mathbf{x}_{i}\) is in \(\mathbb{R}^{n}\). If \(A\) is an \(m \times n\) matrix and \(A \mathbf{x}_{i}=\mathbf{0}\) for each \(i\), show that \(A \mathbf{y}=\mathbf{0}\) for every vector \(\mathbf{y}\) in \(U\)

Short answer

Vectors in the span of \(\{\mathbf{x}_i\}\) map to \(\mathbf{0}\) by \(A\).

Step-by-step solution

Understanding the components

We are given that vectors \(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\) are in \(\mathbb{R}^n\) and they span a subspace \(U\). Also, \(A\) is an \(m \times n\) matrix such that the product of \(A\) with any of these vectors results in the zero vector \(\mathbf{0}\) in \(\mathbb{R}^m\). We need to show this property holds for any vector \(\mathbf{y}\) in the span of these vectors as well.

Defining Vector \( \mathbf{y} \) in \( U \)

Since \( U \) is the span of \(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\), any vector \(\mathbf{y} \in U\) can be expressed as a linear combination: \(\mathbf{y} = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k\), where \( c_1, c_2, \ldots, c_k \) are scalars.

Multiply by matrix \( A \)

Compute the product \( A\mathbf{y} \). Substituting the expression for \(\mathbf{y}\), we have: \[ A \mathbf{y} = A(c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k) \].

Using Matrix Linearity

Apply the distributive property and linearity of matrix multiplication: \[ A \mathbf{y} = c_1 A \mathbf{x}_1 + c_2 A \mathbf{x}_2 + \cdots + c_k A \mathbf{x}_k \].

Substituting \(A \mathbf{x}_i = \mathbf{0} \)

Given that \(A \mathbf{x}_i = \mathbf{0} \) for each \(i\), substitute \(\mathbf{0}\) into the expression: \[ A \mathbf{y} = c_1 \mathbf{0} + c_2 \mathbf{0} + \cdots + c_k \mathbf{0} \].

Final simplification

Since all terms are scalar multiples of the zero vector, they simplify to the zero vector: \[ A \mathbf{y} = \mathbf{0} + \mathbf{0} + \cdots + \mathbf{0} = \mathbf{0} \].

Conclusion

Thus, for any vector \( \mathbf{y} \) in \( U \), it holds that \( A\mathbf{y} = \mathbf{0} \). The property \(A\mathbf{x}_i = \mathbf{0} \) extends to any vector in the span \( U \).

Key concepts

Vector Spaces

In linear algebra, the concept of vector spaces is fundamental. A vector space is a collection of vectors that can be added together and multiplied by scalars to produce another vector within the same space. This is crucial because it provides a framework for many operations that we perform with vectors. In our given problem, the span of vectors \( \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k \) forms a subspace \( U \) within the vector space \( \mathbb{R}^n \).

• **Span:** The span of a set of vectors is all possible linear combinations of those vectors. For instance, any vector \( \mathbf{y} \) in \( U \) can be expressed using the vectors \( \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k \).
• **Subspace:** A subspace is a subset of a vector space that is itself a vector space under the operations of vector addition and scalar multiplication. In this case, \( U \) is a subspace of \( \mathbb{R}^n \).

The operations allowed in a vector space, such as addition and scalar multiplication, are critical in understanding how vectors interact with matrices, leading into our discussion on matrix multiplication.

Matrix Multiplication

Matrix multiplication is a procedure where two matrices are multiplied to produce another matrix. In our problem, we're particularly interested in multiplying the matrix \( A \) by vectors from our space \( U \). Understanding how this operation works is key to solving the exercise.

• **Linearity:** One essential property of matrix multiplication is its linearity. This means that \( A(c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k) \) can be expanded to \( c_1A\mathbf{x}_1 + c_2A\mathbf{x}_2 + \ldots + c_kA\mathbf{x}_k \).
• **Zero Product:** Given the problem, we are told that \( A\mathbf{x}_i = \mathbf{0} \) for each vector \( \mathbf{x}_i \) in the subspace. This simplifies each term of our expanded expression to zero, demonstrating that \( A\mathbf{y} = \mathbf{0} \) for any vector \( \mathbf{y} \) in \( U \).

This demonstrates how efficiently operations on matrices facilitate manipulations and proofs in linear algebra, particularly when validating results across entire subspaces.

Linear Combinations

Linear combinations are expressions that involve adding together multiple vectors, each scaled by a certain factor. This concept is crucial to our understanding of vector spaces and the exercise at hand. When working in a vector space, we are often interested in how vectors can be combined using scalars. This allows us to construct new vectors, like the vector \( \mathbf{y} \) in the span \( U \) from the problem.

• **Building New Vectors:** In vector spaces, any vector can frequently be represented as a linear combination. For the subspace \( U \), any \( \mathbf{y} \) is \( c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k \).
• **Role in Proofs:** In our problem, this representation of vectors as linear combinations allows us to break down the multiplication \( A\mathbf{y} \) step by step, using properties like linearity to conclude the result \( \mathbf{0} \).

Understanding linear combinations allows us to see how vectors interrelate within their space, providing a deeper insight into the structures and results of linear transformations in algebra.