Question

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

Short answer

a. \(\operatorname{im} A = \operatorname{im}(A V)\); b. \(\operatorname{dim}(\operatorname{im} A) = \operatorname{dim}(\operatorname{im}(U A))\).

Step-by-step solution

Understanding Image Transformation with Invertible Matrix

To show that \( \operatorname{im} A = \operatorname{im}(A V) \) for every invertible \( n \times n \) matrix \( V \), consider the effect of \( V \) on \( A \). The image of \( A \), \( \operatorname{im} A \), is the set of all vectors \( A \mathbf{x} \) for every vector \( \mathbf{x} \). Now, the image of \( A V \), \( \operatorname{im}(A V) \), is the set of all vectors \( A(V \mathbf{x}) \). Since \( V \) is invertible, every vector \( \mathbf{x} \) can be expressed as \( V \mathbf{y} \) for some vector \( \mathbf{y} \), and thus \( A V \mathbf{x} = A \mathbf{y} \). Therefore, these two image sets are equivalent, meaning \( \operatorname{im} A = \operatorname{im}(A V) \).

Manipulating the Image with Invertible Matrix

This step confirms that for every invertible \( n \times n \) matrix \( V \), the operation \( A V \) does not change the image of \( A \). Here, any vector \( A \mathbf{x} \) in the image can be achieved by a vector \( A V (V^{-1} \mathbf{x}) \), and conversely, every vector \( A V \mathbf{y} \) can be obtained from some vector in the original space since \( V \) is invertible. Therefore, the transform \( A V \) changes coordinates, not the span or the image.

Analyzing Dimension Equality with Invertible Matrix

To show that \( \operatorname{dim}(\operatorname{im} A)=\operatorname{dim}(\operatorname{im}(U A)) \) for every invertible \( m \times m \) matrix \( U \), consider that the rank of a matrix (dimension of its image) is preserved under pre-multiplication by an invertible matrix. Let \( U \) be such a matrix: \( \operatorname{im}(U A) \) is the set of vectors \( U A \mathbf{x} \). Since \( U \) is invertible, its rank is equal to rank of \( A \). Applying \( U \) does not change the rank because each transformation retains the span due to invertibility.

Constructing a Basis to Compare Dimensions

Assume that \( \left\{ \mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{k} \right\} \) is a basis of \( \operatorname{im}(U A) \). Since \( U \) is invertible, each basis vector \( \mathbf{y}_i \) can be written as \( U \mathbf{z}_i \) for some \( \mathbf{z}_i \). Then \( U \mathbf{z}_i = \, \mathbf{y}_i \), and applying \( U^{-1} \) yields \( \mathbf{z}_i = U^{-1} \mathbf{y}_i \). Thus, \( \left\{U^{-1} \mathbf{y}_{1}, U^{-1} \mathbf{y}_{2}, \ldots, U^{-1} \mathbf{y}_{k} \right\} \) forms a basis for \( \operatorname{im} A \), confirming equal dimension.

Key concepts

Matrix Image

The image of a matrix, often referred to as the matrix image, is an essential concept in linear algebra. When dealing with a matrix \( A \), the image, \( \operatorname{im} A \), consists of all possible outputs when multiplying \( A \) by any vector \( \mathbf{x} \). In other words, it is the set of all vectors that can be expressed in the form \( A\mathbf{x} \). This forms a subspace of the codomain of \( A \).

The matrix image is critical when exploring how transformations affect vector spaces. For instance, if you transform a matrix \( A \) using another matrix \( V \) that is invertible, the overall transformation does not alter the image. Specifically, \( \operatorname{im} A = \operatorname{im}(A V) \) holds since \( V \)'s invertibility ensures that any resulting vector can be traced back to an original vector in the image of \( A \). This property emphasizes the resilient nature of the image under invertible transformations.

Invertible Matrices

Invertible matrices are unique in that they have a two-sided multiplicative inverse, meaning if \( V \) is an invertible matrix, there exists another matrix \( V^{-1} \) such that \( V V^{-1} = V^{-1} V = I \), where \( I \) is the identity matrix.

These matrices play a critical role in linear algebra because applying an invertible matrix to another matrix or vector equates to a reversible transformation. In geometric terms, this means the transformation does not compress, collapse, or otherwise alter the original space in terms of dimension and structure.

• They preserve dimensions: The transformation by an invertible matrix doesn't change the underlying dimension of a vector space.
• Invertible matrices maintain equivalencies: For instance, \( A \) and \( A V \) share the same image, demonstrating that while coordinate representations may shift, the span remains intact.

Understanding how invertible matrices work is crucial for interpreting matrix manipulations and ensuring transformations remain valid and reversible.

Matrix Rank

The rank of a matrix is essentially the dimension of its image, often symbolized as \( \operatorname{rank}(A) \). It tells us how many dimensions the output space of a matrix transformation spans. If you think of a transformation applying to vectors in a space, the rank conveys how much of that space is accessible by the transformation using the given matrix.

For example, if you have a matrix \( A \) and you premultiply it by an invertible square matrix \( U \), \( U A \) maintains the same rank as \( A \). This is due to the fact that an invertible matrix retains the linear independence of columns or rows during transformation, ensuring that no "dimension" of the space is lost. Thus, for any invertible matrix \( U \), \( \operatorname{dim} (\operatorname{im} A) = \operatorname{dim} (\operatorname{im}(U A)) \).

• The rank helps to identify full dimension span potential.
• Consistent rank implies the same dimensional capabilities post-transformation.

Understanding a matrix's rank is vital when determining its ability to act as a full projection of a vector space.

Linear Algebra

Linear algebra is a branch of mathematics centered around vectors, vector spaces, and linear transformations. It plays a foundational role in various scientific fields due to its approaches to solving systems of linear equations, among other applications.

At the heart of linear algebra, you find matrices and their capabilities to transform vector inputs into outputs in possibly different dimensions. Through operations like matrix multiplication, determination of image, rank, and consideration of invertibility, linear algebra provides a structured approach to evaluating transformations.

In educational contexts, linear algebra simplifies understanding complex spaces and transformations:

• Matrix transformations facilitate modeling of real-world problems.
• Concepts such as rank and invertibility guide the interpretation of vector space properties.
• Image analysis assists in grasping the scope of dimensional impact.

Gaining a mastery of linear algebra concepts allows students and professionals alike to leverage these mathematical tools in analysis, engineering, computer science, and beyond.