Question

Let \(A\) be an \(n \times n\) matrix of rank \(r .\) If \(U=\left\\{X\right.\) in \(\left.\mathbf{M}_{n n} \mid A X=0\right\\},\) show that \(\operatorname{dim} U=n(n-r)\).

Short answer

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

Step-by-step solution

Understand the Problem

The problem asks us to find the dimension of the solution space, denoted as \( U \), of the homogenous matrix equation \( AX = 0 \). Here, \( A \) is an \( n \times n \) matrix and \( X \) is an \( n \times n \) matrix. We need to determine how \( \operatorname{dim} U \) relates to \( n \) and the rank of \( A \), \( r \).

Recall a Key Linear Algebra Concept

The rank-nullity theorem states that for a matrix \( A \) of size \( n \times n \), the dimension of the column space (rank) plus the dimension of the null space (nullity) equals \( n \). That is, \( \text{rank}(A) + \text{nullity}(A) = n \). For an \( n \times n \) matrix \( A \) with rank \( r \), the nullity is \( n - r \).

Determine the Structure of Solutions

The solutions \( X \) must satisfy \( AX = 0 \), meaning each column of \( X \) is in the null space of \( A \). Therefore, each column of \( X \) is a vector in an \( n \)-dimensional space with a dimension of \( n - r \) for the null space.

Calculate the Dimension of the Solution Space

Since each column of the \( n \times n \) matrix \( X \) lies in the \( n - r \) dimensional null space, and there are \( n \) such columns in \( X \), the dimension of the space of all such matrices \( X \) is the product \( n \times (n - r) \). Hence, \( \operatorname{dim} U = n(n - r) \).

Key concepts

Rank-Nullity Theorem

The rank-nullity theorem is a fundamental concept in linear algebra. It establishes a relationship between the rank of a matrix and the dimension of its null space. To break it down, the theorem states:

• The rank of a matrix, denoted as \( \text{rank}(A) \), refers to the maximum number of linearly independent column vectors in the matrix.
• The nullity of the matrix, \( \text{nullity}(A) \), is the dimension of the null space of the matrix. It counts the number of linearly independent solutions to the equation \( AX = 0 \).
• The theorem gives us the equation: \( \text{rank}(A) + \text{nullity}(A) = n \), where \( n \) is the number of columns in the matrix.

In our exercise, an \( n \times n \) matrix \( A \) with rank \( r \) implies a nullity of \( n - r \). This directly impacts the dimension of the solution space \( U \). We use this theorem to find that \( \text{dim} U = n(n - r) \). It effectively tells us how failures of columns to be independent (null space dimension) relate to the rank.

Null Space

The null space of a matrix is a vital idea in understanding linear transformations. It represents all vectors that are mapped to the zero vector when multiplied by the matrix. For a matrix \( A \), the null space can be defined as:

• It includes all possible solutions \( X \) to the equation \( AX = 0 \).
• The dimension of this set of solutions is called the nullity of the matrix.
• In scenarios where the null space is larger, meaning a higher nullity, there are more vectors that satisfy the homogenous equation \( AX = 0 \).

In the given exercise, the null space determines the structure of the matrices \( X \) that satisfy \( AX = 0 \). Each column in such a matrix \( X \) must be in this null space. Thus, understanding the concept of null space allows us to compute the dimension of \( U \), which is \( n(n - r) \).

Matrix Dimension

Matrix dimension is a basic but crucial aspect of linear algebra. Every matrix is characterized by its dimensions, which are given by the number of its rows and columns. Let's see why matrix dimensions matter:

• An \( n \times n \) matrix indicates both dimensions, listing the equal number of rows and columns.
• The dimensions dictate operations permissible between matrices, like addition, multiplication, and determining spaces.
• In the context of our exercise, the dimension of the solution space is affected by the dimensions of matrix \( A \), which has \( n \) columns and rows.

For the exercise, understanding that \( A \) has \( n \times n \) structure helps us deduce \( \text{dim} U = n(n - r) \). Each dimension expresses how matrix transformations and solutions are structured, especially in finding the solution spaces associated with given linear equations.