Question

If \(U\) is a subspace of \(\mathbb{R}^{n}\), show how to find an \(n \times n\) matrix \(A\) such that \(U=\\{\mathbf{x} \mid A \mathbf{x}=\mathbf{0}\\} \)

Short answer

Find a basis for \( U \), extend it to \( \mathbb{R}^n \), use additional vectors to construct matrix \( A \).

Step-by-step solution

Identify a Basis for Subspace U

First, identify a basis for the subspace \( U \). Suppose the dimension of \( U \), denoted as \( ext{dim}(U) \), is \( k \), then you need \( k \) linearly independent vectors that span \( U \). Let these vectors be \( extbf{v}_1, extbf{v}_2, extbf{v}_3, extbf{...}, extbf{v}_k \).

Extend Basis to a Basis of Entire Space

To construct our matrix \( A \), we must first extend the basis \( \{ \textbf{v}_1, \textbf{v}_2, \dots, \textbf{v}_k \} \) to a basis for \( \mathbb{R}^n \), which requires \( n \) vectors total (because every basis for \( \mathbb{R}^n \) requires \( n \) linearly independent vectors). Let the additional vectors be \( \textbf{v}_{k+1}, \textbf{v}_{k+2}, \ldots, \textbf{v}_n \).

Construct Matrix with Additional Vectors

Now, construct a matrix \( B \) using the additional vectors \( \textbf{v}_{k+1}, \textbf{v}_{k+2}, \ldots, \textbf{v}_n \), as its rows. Matrix \( B \) will have size \((n-k) \times n\), because there are \( n-k \) vectors we added. The kernel (null space) of this matrix \( B \) will give us the subspace \( U \).

Form the Desired Matrix A

Finally, construct the matrix \( A \) by ensuring it is an \( n \times n \) matrix. You can augment \( B \) with a row of zeros to reach the full \( n \times n \) dimension. The row space of \( A \) is orthogonal to \( U \), meaning \( U = \{ \textbf{x} \mid A \textbf{x} = \textbf{0} \} \).

Key concepts

Matrix Construction

Matrix construction is a fundamental concept in linear algebra. In this exercise, we want to construct an \( n \times n \) matrix \( A \) such that a given subspace \( U \) of \( \mathbb{R}^n \) can be represented as the null space of \( A \). To do this, we follow a structured approach:

• First, we identify a basis for \( U \), which consists of the smallest set of linearly independent vectors that spans \( U \).


• Next, we extend this basis to create a basis for the entire space \( \mathbb{R}^n \). This involves adding more vectors to the old basis until we have exactly \( n \) vectors.


• Using the vectors that were added to form the full basis, we build an initial matrix \( B \). \( B \) will have rows made up of these new basis vectors.


• Finally, we construct our needed \( n \times n \) matrix \( A \) by augmenting \( B \) so that \( A \) is of the proper size, and ensures that its row space is orthogonal to \( U \).

This structured structure ensures that \( U \) is exactly the null space of \( A \), satisfying the condition \( A\mathbf{x} = \mathbf{0} \) for every vector \( \mathbf{x} \) in \( U \). The precision and clarity in these steps are crucial in matrix construction, especially while working within \( \mathbb{R}^n \).

Subspace of \( \mathbb{R}^n \)

A subspace of \( \mathbb{R}^n \) is a special subset of \( \mathbb{R}^n \) that is closed under addition and scalar multiplication. This means that any vector within this subspace, when added to another vector in the same subset or multiplied by a scalar, remains within the subspace.

In our problem, the subspace \( U \) is defined by the property \( U = \{ \mathbf{x} \mid A\mathbf{x} = \mathbf{0} \} \). This condition represents a way of describing \( U \) through a system of linear equations provided by the rows of matrix \( A \).

Understanding subspaces involves recognizing:

• The zero vector is always included in a subspace.
• Any linear combination of vectors in the subspace stays within the subspace.
• Subspaces can be defined using a basis, a set of vectors that span the subspace and are linearly independent.

By defining \( U \) as the null space of \( A \), we are acknowledging that \( U \) includes all solutions to \( A\mathbf{x} = \mathbf{0} \). This elegant connection between linear equations and geometric spaces is central to understanding linear algebra.

Basis Extension

Basis extension is a method used in linear algebra to expand a basis for a subspace into a basis for the entire vector space. When starting with a basis \( \{\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_k\} \) for the subspace \( U \), we need to extend it to a full basis for \( \mathbb{R}^n \).

The process goes as follows:

• Ensure the existing set is linearly independent.
• Select additional vectors \( \mathbf{v}_{k+1}, \ldots, \mathbf{v}_n \) to add to the original basis, ensuring the new set is also linearly independent and spans \( \mathbb{R}^n \).


• The newly included vectors must not lie in the span of the original basis to maintain linear independence.

This method allows us to explore all dimensions of \( \mathbb{R}^n \) while ensuring that we still preserve the properties of the original subspace. This extension is the pivotal step for constructing matrix \( A \), as it provides the necessary framework to include every dimension of the vector space for the matrix representation.

Null Space

The null space of a matrix, also known as the kernel, is a set of all vectors \( \mathbf{x} \) such that \( A\mathbf{x} = \mathbf{0} \). This concept is instrumental in linear algebra as it helps us understand the solutions to homogeneous systems of linear equations.

For matrix \( A \), the null space can be described as:

• Comprising all solutions to the equation \( A\mathbf{x} = \mathbf{0} \).


• Being a subspace itself within \( \mathbb{R}^n \), by definition of how linear transformations act on space.


• Dimensionality, also termed as the nullity, characterizes how many free parameters are in the system.

The connection of matrix \( A \) and its null space aligns very well with the concept of solving for subspaces, as demonstrated. By finding a matrix that represents \( U = \{ \mathbf{x} \mid A \mathbf{x} = \mathbf{0} \} \), we make a geometric representation of the solution space. This highlights how understanding the null space is essential not only for solving linear systems but also for interpreting vector spaces geometrically.