Problem 15 - Differential Equations and Linear Algebra

Question

Show that a \(6 \times 4\) matrix \(A\) with nullity \((A)=0\) must have rowspace \((A)=\mathbb{R}^{4} .\) Is colspace \((A)=\mathbb{R}^{4} ?\)

Step-by-Step Solution

Answer

A \(6 \times 4\) matrix \(A\) with nullity\((A)=0\) has rank(A) = 4, implying that the row space\((A) = \mathbb{R}^4\). However, the column space\((A)\) is a subspace of \(\mathbb{R}^6\) and cannot be equal to \(\mathbb{R}^4\).

Step 1: Define the given terms

First, let's define what the given terms mean. Nullity\((A)\): The nullity of a matrix A is the dimension of the null space of A, i.e., the number of linearly independent solutions to the equation Ax = 0. Row space\((A)\): The row space of a matrix A is the space spanned by its row vectors. Column space\((A)\): The column space of a matrix A is the space spanned by its column vectors.

Step 2: Understand the problem in terms of the rank-nullity theorem

We need to apply the rank-nullity theorem, which states that for an \(m \times n\) matrix A, the sum of rank(A) and nullity(A) always equals n. Here, rank(A) = dimension of row space(A), and nullity(A) = dimension of null space(A). The given matrix A is a \(6 \times 4\) matrix with nullity(A) = 0.

Step 3: Deduce the rank(A)

Since the nullity(A) = 0, applying the rank-nullity theorem to the given matrix A, we have: rank(A) + nullity(A) = n rank(A) + 0 = 4 rank(A) = 4

Step 4: Conclude row space(A)

Since rank(A) is 4, this means that the row space of A has 4 linearly independent row vectors, which span a space of dimension 4. So, the row space\((A) = \mathbb{R}^4\).

Step 5: Discuss column space(A)

The column space (also called the range) of a matrix is the space spanned by its column vectors. Since the given matrix \(A\) is of size \(6\times 4\), it has 4 column vectors in \(\mathbb{R}^6\), not in \(\mathbb{R}^4\). Hence, the column space\((A)\) is a subspace of \(\mathbb{R}^6\) and cannot be equal to \(\mathbb{R}^4\). To summarize: We showed that a \(6 \times 4\) matrix \(A\) with nullity\((A)=0\) must have row space \((A)=\mathbb{R}^4\). However, the column space\((A)\) cannot be equal to \(\mathbb{R}^4\) since it is a subspace of \(\mathbb{R}^6\) instead.

Key Concepts

Nullity of a Matrix

The concept of nullity is integral to understanding the structure of linear equations and the matrices that represent them. The nullity of a matrix, often denoted as nullity(A), is the dimension of the null space of A. In other words, it is the number of linearly independent solutions to the homogeneous equation Ax = 0, where A is the matrix in question and x is a vector of variables.

The nullity provides insight into the number of parameters one can freely choose when solving a system of linear equations. If nullity(A) is zero, this implies that there are no free variables and all solutions to Ax = 0 are trivial – meaning, x must be the zero vector. Hence, there are no directions in which you can move from the origin without violating the equation.

A practical way to determine the nullity is through row reduction to echelon form and identifying the pivot columns. The total number of columns minus the number of pivot columns gives us the nullity. A nullity of zero, as in the original exercise, indicates that all columns have pivots and thus contribute to the unique solution of the equation.

Row Space of a Matrix

The row space of a matrix is another fundamental concept in linear algebra, which can sometimes seem enigmatic at first glance. Simply put, the row space of matrix A, denoted as row space(A), is the set of all possible linear combinations of its row vectors.

Understanding the row space of a matrix helps in determining the rank of a matrix, which is the maximum number of linearly independent rows. The rank-nullity theorem connects the row space directly to the nullity, as it states that the rank of the matrix plus its nullity will equal the total number of columns, symbolically shown as rank(A) + nullity(A) = n.

In terms of the original exercise, since the matrix A had a nullity of zero, all rows were linearly independent, and thus, the row space spanned \(\mathbb{R}^{4}\), implying that any vector in that space could be expressed as a combination of the rows of A.

Column Space of a Matrix

To fully grasp the concept of the column space of a matrix, denoted as col space(A), one needs to picture it as the span of all the column vectors within the matrix A. Intuitively, it represents all the linear combinations of the column vectors, similar to how row space is considered for row vectors.

The dimension of the column space is directly tied to the rank of the matrix, as both concepts describe the number of linearly independent vectors; for the column space, it's the columns of A. The rank is also the dimension of the image or range of the matrix transformation.

Bringing this back to the original problem, we concluded that the column space of a 6x4 matrix A exists in \(\mathbb{R}^6\) because it is spanned by four column vectors that stretch into a six-dimensional space. Hence the column space can't equal \(\mathbb{R}^4\), as each of the column vectors adds another dimension to the subspace of \(\mathbb{R}^6\). This clarification is essential for distinguishing between the concepts of row space and column space, which students often conflate or misunderstand.