Question

Fix a column \(\mathbf{y} \neq \mathbf{0}\) in \(\mathbb{R}^{n}\) and let \(U=\left\\{A\right.\) in \(\left.\mathbf{M}_{n n} \mid A \mathbf{y}=\mathbf{0}\right\\} .\) Show that \(\operatorname{dim} U=n(n-1)\)

Short answer

The dimension of \( U \) is \( n(n-1) \).

Step-by-step solution

Identify the Condition for A

We are given that a matrix \( A \in U \) satisfies the condition \( A\mathbf{y} = \mathbf{0} \), where \( \mathbf{y} \) is a fixed non-zero vector in \( \mathbb{R}^n \). Therefore, the columns of \( A \) must lie in the null space of the linear transformation defined by \( \mathbf{y}^T \). This transformation maps \( \mathbf{x} \in \mathbb{R}^n \) to \( \mathbf{y}^T \mathbf{x} \).

Determine the Nullity of \( y^T \)

The map \( \mathbf{y}^T \) is a 1-dimensional row vector, transforming \( \mathbf{x} \) to a scalar. The null space of this map consists of all vectors orthogonal to \( \mathbf{y} \). Since \( \mathbf{y} eq \mathbf{0} \), the null space has dimension \( n-1 \).

Matrix \( A \) from Null Space Vectors

Each column of the matrix \( A \) can independently be any vector from the null space of \( \mathbf{y}^T \). Given that the null space is \( n-1 \) dimensional, each column has \( n-1 \) degrees of freedom to choose vectors orthogonal to \( \mathbf{y} \).

Compute Dimensionality of Matrix Set U

There are \( n \) columns in the matrix \( A \), each with \( n-1 \) degrees of freedom because they are chosen from the \( n-1 \) dimensional null space. Thus, the degrees of freedom for \( A \) is \( n(n-1) \), leading to the dimensionality of the space \( U \) being \( n(n-1) \).

Confirm Dimensionality

Since each column vector in \( A \) can be selected independently from a \( n-1 \) dimensional space, and there are \( n \) such columns, the dimension of \( U \) is correctly determined as \( n(n-1) \).

Key concepts

Matrix Null Space

In linear algebra, the null space of a matrix, also known as the kernel, is a fundamental concept that involves finding all the vectors that transform to the zero vector under a given matrix operation. For a linear transformation represented by a matrix like \( \mathbf{y}^T \) in the given exercise, the null space is comprised of all vectors \( \mathbf{x} \) such that \( \mathbf{y}^T \mathbf{x} = 0 \). This means that vectors in the null space are orthogonal to the vector \( \mathbf{y} \).

Some important aspects of the null space include:

• It captures the solutions to the homogeneous equation \( A\mathbf{x} = \mathbf{0} \).
• Its dimension is known as the nullity of the matrix.
• The nullity tells us the degree of freedom within the transformation that maps vectors to zero.

In this exercise, the matrix \( A \) is chosen such that each of its columns lies within the null space of \( \mathbf{y}^T \). Since the vector \( \mathbf{y} \) is non-zero, its related null space in \( \mathbb{R}^n \) has a dimensionality of \( n-1 \). This means there is a rich set of vectors, besides the zero vector, that can form the columns of \( A \).

Matrix Dimensionality

Understanding the dimensionality of matrices is crucial in solving many linear algebra problems. The dimension of a matrix set like \( U \), in this context, refers to the count of independent parameters or vectors needed to span that space. In this exercise, we are working with the idea of calculating the dimension of a specific set of matrices where each matrix satisfies the condition \( A\mathbf{y} = \mathbf{0} \).

Here's how we approach determining this dimensionality:

• The null space associated with \( \mathbf{y}^T \) has \( n-1 \) dimensions since it contains all vectors orthogonal to \( \mathbf{y} \).
• Each column of the matrix \( A \) can independently take any value from the null space, offering \( n-1 \) degrees of freedom per column.
• Given there are \( n \) columns, the total dimensionality of the matrices satisfying the condition in \( U \) is calculated by multiplying \( n \times (n-1) \), which simplifies to \( n(n-1) \).

This determination is critical as it helps us understand the vast space or freedom we have in constructing such matrices, providing insights into both the structure of matrices and applications like solutions to systems of equations.

Linear Transformations

Linear transformations are a key part of understanding how matrices function in linear algebra. A linear transformation is a mathematical operation that maps one vector space to another while preserving the operations of vector addition and scalar multiplication.

Key aspects to grasp about linear transformations include:

• A matrix can represent a linear transformation by mapping vectors from one space into another: for this exercise, \( \mathbf{y}^T \) maps from \( \mathbb{R}^n \) to the scalar field \( \mathbb{R} \).
• The null space of a transformation provides crucial insight into which elements are mapped to zero, reflecting on the transformation's kernel.
• Understanding transformations aids in modeling various phenomena in science and engineering, where matrix operations describe systems or data transformations.

In the case of this exercise, the matrix \( A \) embodies a transformation wherein every column, having \( n-1 \) degrees of freedom, is effectively a vector in the null space of the transformation defined by \( \mathbf{y}^T \). This reaffirms the interconnectedness of matrix operations and vector spaces in defining and determining the effects of transformations.