Question

Let \(A\) denote an \(m \times n\) matrix. a. Show that null \(A=\) null \((U A)\) for every invertible \(m \times m\) matrix \(U\) b. Show that \(\operatorname{dim}(\operatorname{null} A)=\operatorname{dim}(\) null \((A V))\) for every invertible \(n \times n\) matrix \(V\). [Hint: If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is a basis of null \(A,\) show that \(\left\\{V^{-1} \mathbf{x}_{1}, V^{-1} \mathbf{x}_{2}, \ldots, V^{-1} \mathbf{x}_{k}\right\\}\) is a basis of \(\operatorname{null}(A V) .]\)

Short answer

For any invertible matrices \(U\) and \(V\), \(\text{null}(A) = \text{null}(UA)\) and \(\operatorname{dim}(\text{null}(A)) = \operatorname{dim}(\text{null}(AV))\).

Step-by-step solution

Understand the Null Space Definition

The null space (or kernel) of a matrix \(A\), denoted as \(\text{null}(A)\), is the set of all vectors \(\mathbf{x}\) such that \(A\mathbf{x} = \mathbf{0}\). Our goal is to analyze how this set changes when \(A\) is multiplied by an invertible matrix \(U\) or \(V\).

Show null \(A = \text{null}(UA)\)

To show \(\text{null}(A) = \text{null}(UA)\), consider that a vector \(\mathbf{x}\) is in \(\text{null}(A)\) if and only if \(A \mathbf{x} = \mathbf{0}\). For \(\text{null}(U A)\), we have \(U A \mathbf{x} = \mathbf{0}\). Because \(U\) is invertible, the only solution to \(U A \mathbf{x} = \mathbf{0}\) is if \(A \mathbf{x} = \mathbf{0}\). Thus, any \(\mathbf{x}\) satisfying the equation for \(\text{null}(UA)\) must also satisfy it for \(\text{null}(A)\), and vice versa.

Analyze the Basis of \(\text{null}(A)\) and \(\text{null}(AV)\)

Let \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\}\) be a basis of \(\text{null}(A)\), meaning \(A\mathbf{x}_i = \mathbf{0}\) for all \(i\). We need to show \(\{V^{-1}\mathbf{x}_1, V^{-1}\mathbf{x}_2, \ldots, V^{-1}\mathbf{x}_k\}\) is a basis for \(\text{null}(AV)\), which occurs if \((AV)(V^{-1} \mathbf{x}_i) = A \mathbf{x}_i = \mathbf{0}\). Hence, the vectors \(V^{-1} \mathbf{x}_i\) span \(\text{null}(AV)\) without changing its dimensionality.

Conclusion on Dimensions

The transformation from \(\text{null}(A)\) to \(\text{null}(AV)\) involves a change of basis using the invertible matrix \(V^{-1}\), which does not affect the dimension. Therefore, \(\operatorname{dim}(\text{null}(A)) = \operatorname{dim}(\text{null}(AV))\).

Key concepts

Invertible Matrix

An invertible matrix, also known as a non-singular or non-degenerate matrix, is a square matrix that possesses an inverse. A matrix \( U \) is invertible if there exists another matrix \( U^{-1} \) such that the product of \( U \) and \( U^{-1} \) is an identity matrix, i.e., \( U U^{-1} = U^{-1} U = I \).

The properties of invertible matrices play a crucial role in linear algebra and are vital for understanding the behavior of matrix transformations. One essential property is that multiplying an invertible matrix by another matrix does not change the null space of the original matrix. This is because the presence of the invertible matrix is effectively "canceled out" when solving the homogeneous equation.

• For example, if \( A \) is an \( m \times n \) matrix, and \( U \) is an \( m \times m \) invertible matrix, then the null space remains the same, i.e., \( \text{null}(A) = \text{null}(UA) \).
• Similarly, if \( V \) is an \( n \times n \) invertible matrix, the null space affected by right multiplication also retains its dimension despite the transformation.

In mathematical analysis, understanding invertible matrices helps solve systems of linear equations effectively and ensures transformations preserve essential properties such as dimensionality.

Matrix Dimensions

The dimensions of a matrix are given in terms of the number of its rows and columns, typically represented as \( m \times n \). These dimensions are a fundamental aspect of matrix algebra as they define the size and shape of the matrix and dictate the conditions for performing operations such as matrix multiplication.

In the context of null spaces and invertible matrices, dimensions interact with the concept of rank and nullity. These two properties are related through the Rank-Nullity Theorem, a key principle stating that for an \( m \times n \) matrix \( A \), the sum of the rank (number of linearly independent columns or rows) and the nullity (dimension of the null space) equals the number of columns \( n \). This can be expressed as: \[ \text{rank}(A) + \text{nullity}(A) = n. \]

• The dimension of the null space, known as nullity, represents how many solutions exist to the equation \( A \mathbf{x} = \mathbf{0} \) other than the trivial solution.
• Understanding these dimensions is crucial when analyzing how operations with invertible matrices alter or maintain properties of the matrices involved, particularly when dealing with matrix equations or linear transformations.

Thus, when multiplying a matrix by an invertible matrix, either from the left or right, the overall dimensionality concerning its null space remains unaffected.

Basis of Null Space

The basis of the null space of a matrix is defined as a set of linearly independent vectors that span this null space. In linear algebra, the matrix's null space refers to all solutions \( \mathbf{x} \) of the linear system \( A\mathbf{x} = \mathbf{0} \). The dimension of this subspace is the nullity of the matrix.

To form a basis for the null space, vectors within this space must satisfy certain conditions:

• They must be linearly independent, meaning that no vector in the set can be written as a linear combination of the others.
• The set must span the null space, which means any vector from the null space can be expressed as a linear combination of these basis vectors.

An important consideration in understanding matrix operations is realizing that the transformation of a matrix by an invertible matrix, particularly multiplying by an inverted matrix, affects the basis but not its dimension. For instance, if \( \{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\} \) are basis vectors for \( \text{null}(A) \), then the vectors \( \{V^{-1} \mathbf{x}_1, V^{-1} \mathbf{x}_2, \ldots, V^{-1} \mathbf{x}_k\} \) form a basis for \( \text{null}(AV) \).

By applying invertible transformations, the structure and span of the null space basis might alter, but its dimension remains the same, reflecting that fundamental properties are preserved under these operations.