Ab Transpose Is Equal To B Transpose A Transpose

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Kalali

Jun 07, 2025 · 3 min read

Ab Transpose Is Equal To B Transpose A Transpose
Ab Transpose Is Equal To B Transpose A Transpose

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    Proving (AB)ᵀ = BᵀAᵀ: A Deep Dive into Matrix Transpose Properties

    This article will demonstrate the fundamental property of matrix transposition: (AB)ᵀ = BᵀAᵀ. This isn't just a mathematical curiosity; understanding this property is crucial for various linear algebra applications, including machine learning, computer graphics, and physics. We'll explore the proof and delve into its implications.

    Understanding the Basics: Matrix Transpose

    Before diving into the proof, let's quickly recap what a matrix transpose is. The transpose of a matrix, denoted by a superscript 'ᵀ', is a new matrix formed by interchanging the rows and columns of the original matrix. For instance, if we have matrix A:

    A =  [ a b ]
         [ c d ]
    

    Then its transpose, Aᵀ, is:

    Aᵀ = [ a c ]
         [ b d ]
    

    Proving (AB)ᵀ = BᵀAᵀ: A Step-by-Step Approach

    The proof hinges on understanding the elements of the resulting matrices. Let's consider matrices A (m x n) and B (n x p). Their product, AB, will be an (m x p) matrix. We'll analyze the elements of both (AB)ᵀ and BᵀAᵀ to show they are identical.

    1. Element-wise examination of (AB)ᵀ:

    Let's denote the element in the i-th row and j-th column of matrix AB as (AB)ᵢⱼ. This element is calculated as the dot product of the i-th row of A and the j-th column of B:

    (AB)ᵢⱼ = Σₖ Aᵢₖ * Bₖⱼ (where the summation is from k=1 to n)

    Now, the transpose (AB)ᵀ will swap rows and columns. Thus, the element in the i-th row and j-th column of (AB)ᵀ will be the element in the j-th row and i-th column of AB:

    ((AB)ᵀ)ᵢⱼ = (AB)ⱼᵢ = Σₖ Aⱼₖ * Bₖᵢ

    1. Element-wise examination of BᵀAᵀ:

    Let's examine the element in the i-th row and j-th column of BᵀAᵀ. Remember that Bᵀ is a (p x n) matrix and Aᵀ is an (n x m) matrix. Therefore:

    (BᵀAᵀ)ᵢⱼ = Σₖ (Bᵀ)ᵢₖ * (Aᵀ)ₖⱼ

    Since (Bᵀ)ᵢₖ = Bₖᵢ and (Aᵀ)ₖⱼ = Aⱼₖ, we can rewrite this as:

    (BᵀAᵀ)ᵢⱼ = Σₖ Bₖᵢ * Aⱼₖ = Σₖ Aⱼₖ * Bₖᵢ

    The Conclusion:

    Observe that ((AB)ᵀ)ᵢⱼ = (BᵀAᵀ)ᵢⱼ for all i and j. Since the corresponding elements of both matrices are equal, we've conclusively shown that:

    (AB)ᵀ = BᵀAᵀ

    Implications and Applications

    This seemingly simple property has wide-ranging implications. For instance, it simplifies calculations involving transposes of matrix products, which are common in many fields. Understanding this property is essential for:

    • Linear Transformations: Analyzing the effects of successive transformations represented by matrices.
    • Machine Learning: Optimizing algorithms and simplifying computations in gradient descent and other optimization techniques involving matrix operations, such as in calculating the transpose of the Jacobian matrix.
    • Computer Graphics: Efficiently handling rotations and other transformations in 3D graphics.
    • Quantum Mechanics: Simplifying calculations involving quantum operators.

    This proof demonstrates the elegance and power of linear algebra. Mastering this fundamental property is a cornerstone of understanding and applying matrix mathematics effectively. Further exploration into other matrix properties will build upon this foundation.

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