\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
deepdream of a sidewalk
Show Question
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\groupAdd}[1] { +_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \require{mathtools} \)
Math and science::Algebra

Transpose characterized

Instead of thinking of the matrix transpose as the swapping of rows and columns, there is a more meaningful characterization:

Matrix transpose

Let \( A \) be an \( m \times n \) matrix and \( x \in \mathbb{R}^n \) and \( y \in \mathbb{R}^m \) be column vectors. Then the transpose of \( A \), denoted \( A^{T} \) is the matrix such that the following holds:

\[ (Ax)^{T}y = x^{T}(A^{T}y) \]

Example

\[ A = \begin{bmatrix} 7 & 0 \\ 1 & -2 \\ 1 & -2 \end{bmatrix}, \quad x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \quad y = \begin{bmatrix} 4 \\ -1 \end{bmatrix}, \quad y = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]

In \( (Ax)^{T}y \), each elements of \( x \) has global scope in terms of the multiplication:

\[ (Ax)^{T}y = (4) \times \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \times \begin{bmatrix} 7 \\ 1 \\ 1 \end{bmatrix} + (-1) \times \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \times \begin{bmatrix} 0 \\ -2 \\ -2 \end{bmatrix} \]

So each element of \( x \) multiplies by all elements of \( y \).

The two possible orders of multiplication are:

  1. Convert \( x \) to a single vector in the same space as \( y \), then do a dot product.
  2. Carry out two dot products with \( y \), one for each of the vectors that \( x \) will scale, then mix the result based on the elements of \( x \).

With the first order, the \( 3 \times 2 \) intermediate result are two vectors in the same space as \( y \), transposed to row vectors. With the second order, the \( 2 \times 1 \) intermediate result are two scalars representing completed dot products with \( y \).