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Subsection 2.1 The Transpose of a matrix

Definition 2.1.33. Transpose.

The transpose of a matrix \(\boldsymbol{A}, \) denoted \(\boldsymbol{A}^{\mathsf{T}}\text{,}\) is the matrix whose columns are the rows of the given matrix \(\boldsymbol{A}.\)
Find the transpose of the matrix,
\begin{equation*} \boldsymbol{A}_{4\times 3} = \left[ \begin{array}{rrr} 4 \amp -1 \amp 2 \\ 0 \amp 3 \amp -4 \\ 1 \amp -2 \amp 2 \\ 2 \amp 0 \amp 3 \end{array}\right]. \end{equation*}
The transpose is
\begin{equation*} \left(\boldsymbol{A}^{\mathsf{T}}\right)_{3 \times 4} = \left[ \begin{array}{rrrr} 4 \amp 0 \amp 1 \amp 2 \\ -1 \amp 3 \amp -2 \amp 0 \\ 2 \amp -4 \amp 2 \amp 3\\ \end{array}\right]. \end{equation*}
Find the transpose of the given matrix.
\begin{equation*} \boldsymbol{A} = \begin{array}{lrrrr} \lceil \amp {-7} \amp {3} \amp {8} \amp \rceil \\ \lfloor \amp {-2} \amp {5} \amp {7} \amp \rfloor \\ \end{array} \end{equation*}
\(\lceil\) \(\rceil\)
\(\boldsymbol{A}^{\mathsf{T}}=\) \(\vert\) \(\vert\)
\(\lfloor\) \(\rfloor\)
\begin{equation*} \, \end{equation*}
Answer 1.
\(-7\)
Answer 2.
\(-2\)
Answer 3.
\(3\)
Answer 4.
\(5\)
Answer 5.
\(8\)
Answer 6.
\(7\)
Solution.
\begin{equation*} \boldsymbol{A}^{\mathsf{T}} = \begin{array}{lrrr} \lceil \amp {-7} \amp {-2} \amp \rceil \\ \vert \amp {3} \amp {5} \amp \vert \\ \lfloor \amp {8} \amp {7} \amp \rfloor \\ \end{array} \end{equation*}

Subsubsection 2.1.1 Properties of matrix transposes

  • Transpose of a sum.
    The transpose of a sum is equal to the sum of the transposes.
    \begin{equation*} \left(\boldsymbol{A} + \boldsymbol{B}\right)^{\mathsf{T}} = \boldsymbol{A}^{\mathsf{T}} + \boldsymbol{B}^{\mathsf{T}} \end{equation*}
    \(\,\)
    Find \(\left(\boldsymbol{A} + \boldsymbol{B} \right)^{\mathsf{T}} \) both by find the sum first and then the transpose and by finding the individual transposes first.
    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrrr} 1 \amp 3 \amp -1 \amp 0\\ 2 \amp 5 \amp -1 \amp -2 \end{array}\right], \hspace{1cm} \boldsymbol{B} = \left[ \begin{array}{rrrr} 1 \amp 0 \amp -3 \amp 1\\ 0 \amp 3 \amp 2 \amp 0 \end{array}\right]. \end{equation*}
    \(\)
    • Finding sum first
      \begin{equation*} \begin{array}{ccc} \boldsymbol{A} + \boldsymbol{B} \amp =\amp \left[ \begin{array}{rrrr} 1 \amp 3 \amp -1 \amp 0\\ 2 \amp 5 \amp -1 \amp -2 \end{array}\right] + \left[ \begin{array}{rrrr} 1 \amp 0 \amp -3 \amp 1\\ 0 \amp 3 \amp 2 \amp 0 \end{array}\right]\\ \amp = \amp \left[ \begin{array}{rrrr} 2 \amp 3 \amp -4 \amp 1\\ 2 \amp 8 \amp 1 \amp -2 \end{array}\right]\\ \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \left( \boldsymbol{A} + \boldsymbol{B}\right)^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rrrr} 2 \amp 3 \amp -4 \amp 1\\ 2 \amp 8 \amp 1 \amp -2 \end{array}\right]\right)^{\mathsf{T}}\\ \amp = \amp \left[ \begin{array}{rr} 2 \amp 2\\ 3 \amp 8\\ -4 \amp 1\\ 1 \amp -2 \end{array}\right].\\ \end{array} \end{equation*}
    • Finding transposes first
      \begin{equation*} \begin{array}{ccc} \boldsymbol{A}^{\mathsf{T}} \amp =\amp \left(\left[ \begin{array}{rrrr} 1 \amp 3 \amp -1 \amp 0\\ 2 \amp 5 \amp -1 \amp -2 \end{array}\right]\right)^{\mathsf{T}} \\ \amp = \amp \left[ \begin{array}{rr} 1 \amp 2\\ 3 \amp 5\\ -1 \amp -1\\ 0 \amp -2 \end{array}\right]\\ \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \boldsymbol{B}^{\mathsf{T}} \amp =\amp \left( \left[ \begin{array}{rrrr} 1 \amp 0 \amp -3 \amp 1\\ 0 \amp 3 \amp 2 \amp 0 \end{array}\right]\right)^{\mathsf{T}} \\ \amp = \amp \left[ \begin{array}{rr} 1 \amp 0\\ 0 \amp 3\\ -3 \amp 2\\ 1 \amp 0 \end{array}\right]\\ \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \left( \boldsymbol{A} + \boldsymbol{B}\right)^{\mathsf{T}} \amp = \amp \boldsymbol{A}^{\mathsf{T}} + \boldsymbol{B}^{\mathsf{T}} \\ \amp = \amp \left[ \begin{array}{rr} 1 \amp 2\\ 3 \amp 5\\ -1 \amp -1\\ 0 \amp -2 \end{array}\right] + \left[ \begin{array}{rr} 1 \amp 0\\ 0 \amp 3\\ -3 \amp 2\\ 1 \amp 0 \end{array}\right]\\ \amp = \amp \left[ \begin{array}{rr} 2 \amp 2\\ 3 \amp 8\\ -4 \amp 1\\ 1 \amp -2 \end{array}\right]. \end{array} \end{equation*}
    \(\,\)
  • Transpose of a matrix - scalar product.
    The transpose of the product between a matrix and a scalar is equal to the product between the scalar and the transpose of the matrix.
    \begin{equation*} \left(c \boldsymbol{A} \right)^{\mathsf{T}} = c\,\boldsymbol{A}^{\mathsf{T}} \end{equation*}
    \(\,\)
    Find \(\left(3\, \boldsymbol{A}\right)^{\mathsf{T}} \) both by finding the product first and by finding the transpose first.
    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrr} 1 \amp 3 \amp -1 \\ 0 \amp 2 \amp 1 \\ -3 \amp 1 \amp -2 \end{array}\right] \end{equation*}
    • Finding product first
      \begin{equation*} \begin{array}{ccc} 3\, \boldsymbol{A} \amp = \amp 3 \cdot\left[ \begin{array}{rrr} 1 \amp 3 \amp -1 \\ 0 \amp 2 \amp 1 \\ -3 \amp 1 \amp -2 \end{array}\right]\\ \amp = \amp \left[ \begin{array}{rrr} 3 \amp 9 \amp -3 \\ 0 \amp 6 \amp 3 \\ -9 \amp 3 \amp -6 \end{array}\right] \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \left(3\, \boldsymbol{A}\right)^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rrr} 3 \amp 9 \amp -3 \\ 0 \amp 6 \amp 3 \\ -9 \amp 3 \amp -6 \end{array}\right]\right)^{\mathsf{T}}\\ \amp = \amp \left[ \begin{array}{rrr} 3 \amp 0 \amp -9 \\ 9 \amp 6 \amp 3 \\ -3 \amp 3 \amp -6 \end{array}\right].\\ \end{array} \end{equation*}
    • Finding transpose first
      \begin{equation*} \begin{array}{ccc} \boldsymbol{A}^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rrr} 1 \amp 3 \amp -1 \\ 0 \amp 2 \amp 1 \\ -3 \amp 1 \amp -2 \end{array}\right]\right)^{\mathsf{T}}\\ \amp = \amp \left[ \begin{array}{rrr} 1 \amp 0 \amp -3 \\ 3 \amp 2 \amp 1 \\ -1 \amp 1 \amp -2 \end{array}\right] \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \left(3\, \boldsymbol{A}\right)^{\mathsf{T}} = 3\,\boldsymbol{A}^{\mathsf{T}} \amp = \amp 3 \cdot \left[ \begin{array}{rrr} 1 \amp 0 \amp -3 \\ 3 \amp 2 \amp 1 \\ -1 \amp 1 \amp -2 \end{array}\right]\\ \amp = \amp \left[ \begin{array}{rrr} 3 \amp 0 \amp -9 \\ 9 \amp 6 \amp 3 \\ -3 \amp 3 \amp -6 \end{array}\right].\\ \end{array} \end{equation*}
    \(\,\)
  • Transpose of a matrtix - matrix product.
    The transpose of the product of two matrices is equal to the product between the transpose of each matrix in reverse order.
    \begin{equation*} \left( \boldsymbol{A}\, \boldsymbol{B} \right)^{\mathsf{T}} = \boldsymbol{B}^{\mathsf{T}}\,\boldsymbol{A}^{\mathsf{T}} \end{equation*}
    \(\,\)
    Find \(\left( \boldsymbol{A}\, \boldsymbol{B} \right)^{\mathsf{T}}\) first by finding the product and then the transpose, and second by finding the transposes and then the product.
    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rr} 1 \amp 3 \\ 1 \amp 2 \\ 0 \amp -1 \end{array}\right], \hspace{1cm} \boldsymbol{B} = \left[ \begin{array}{rrr} 1 \amp 0 \amp -2 \\ -1 \amp 1 \amp 2 \end{array}\right]. \end{equation*}
    • Finding the product first
      \begin{equation*} \begin{array}{ccc} \boldsymbol{A} \cdot \boldsymbol{B} \amp = \amp \left[ \begin{array}{rr} 1 \amp 3 \\ 1 \amp 2 \\ 0 \amp -1 \end{array}\right]\cdot \left[ \begin{array}{rrr} 1 \amp 0 \amp -2 \\ -1 \amp 1 \amp 2 \end{array}\right]\\ \amp = \amp \left[ \begin{array}{rrr} -2 \amp 3 \amp 4\\ -1 \amp 2 \amp 2\\ 1 \amp -1 \amp -2 \end{array}\right]\\ \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \left(\boldsymbol{A} \cdot \boldsymbol{B}\right)^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rrr} -2 \amp 3 \amp 4\\ -1 \amp 2 \amp 2\\ 1 \amp -1 \amp -2 \end{array}\right]\right)^{\mathsf{T}} \\ \amp = \amp \left[ \begin{array}{rrr} -2 \amp -1 \amp 1\\ 3 \amp 2 \amp -1\\ 4 \amp 2 \amp -2 \end{array}\right]. \end{array} \end{equation*}
    • Finding transposes first
      \begin{equation*} \begin{array}{ccc} \boldsymbol{A}^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rr} 1 \amp 3 \\ 1 \amp 2 \\ 0 \amp -1 \end{array}\right]\right)^{\mathsf{T}} = \left[ \begin{array}{rrr} 1 \amp 1 \amp 0\\ 3 \amp 2 \amp -1 \end{array}\right]\\ \boldsymbol{B}^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rrr} 1 \amp 0 \amp -2 \\ -1 \amp 1 \amp 2 \end{array}\right]\right)^{\mathsf{T}}= \left[ \begin{array}{rr} 1 \amp -1\\ 0 \amp 1 \\ -2 \amp 2 \end{array}\right]\\ \end{array} \end{equation*}
      \begin{equation*} \begin{array}{ccc} \left(\boldsymbol{A} \cdot \boldsymbol{B}\right)^{\mathsf{T}} =\boldsymbol{B}^{\mathsf{T}} \cdot \boldsymbol{A}^{\mathsf{T}} \amp = \amp \left[ \begin{array}{rr} 1 \amp -1\\ 0 \amp 1 \\ -2 \amp 2 \end{array}\right] \cdot \left[ \begin{array}{rrr} 1 \amp 1 \amp 0\\ 3 \amp 2 \amp -1 \end{array}\right]\\ \amp = \amp \left[ \begin{array}{rrr} -2 \amp -1 \amp 1\\ 3 \amp 2 \amp -1\\ 4 \amp 2 \amp -2 \end{array}\right]. \end{array} \end{equation*}
    \(\,\)
  • Transpose of a transpose.
    The transpose of the transpose of a given matrix \(\boldsymbol{A} \) is the original matrix \(\boldsymbol{A} \text{.}\)
    \begin{equation*} \left( \boldsymbol{A}^{\mathsf{T}}\right)^{\mathsf{T}} = \boldsymbol{A} \end{equation*}
    \(\,\)
    For the given matrix find \(\left(\boldsymbol{A}^{\mathsf{T}}\right)^{\mathsf{T}} \text{.}\)
    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 1 \end{array}\right]. \end{equation*}
    \(\)
    \begin{equation*} \begin{array}{ccc} \boldsymbol{A}^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 1 \end{array}\right]\right)^{\mathsf{T}} = \left[ \begin{array}{rr} 1 \amp 1 \\ -1 \amp 1 \end{array}\right],\\ \amp \amp\\ \left(\boldsymbol{A}^{\mathsf{T}}\right)^{\mathsf{T}} \amp = \amp \left(\left[ \begin{array}{rr} 1 \amp 1 \\ -1 \amp 1 \end{array}\right] \right)^{\mathsf{T}} = \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 1 \end{array}\right]. \end{array} \end{equation*}
    \(\,\)
Consider the matrices
\begin{equation*} \boldsymbol{A} = \begin{array}{lrrr} \lceil \amp {-3} \amp {-2} \amp \rceil\\ \lfloor \amp {-1} \amp {-3} \amp \rfloor \end{array}, \hspace{0.5cm} \boldsymbol{B} = \begin{array}{lrrr} \lceil \amp {2} \amp {-1} \amp \rceil\\ \lfloor \amp {-1} \amp {4} \amp \rfloor \end{array},\hspace{0.5cm} \boldsymbol{C} = \begin{array}{lrrr} \lceil \amp {-3} \amp {-3} \amp \rceil\\ \lfloor \amp {1} \amp {-4} \amp \rfloor \end{array}. \end{equation*}
\(\) Find \(\boldsymbol{D}^{\mathsf{T}}\) where
\begin{equation*} \boldsymbol{D} = {4}\,\boldsymbol{B} - {2}\,\boldsymbol{C} + 2\,\boldsymbol{A}\,\boldsymbol{B} - \boldsymbol{A}\,\boldsymbol{C} \end{equation*}
\(\,\)
\(\lceil\) \(\rceil\)
\(\boldsymbol{D}^{\mathsf{T}}=\) \(\lfloor\) \(\rfloor\)
\begin{equation*} \, \end{equation*}
Answer 1.
\(-1\)
Answer 2.
\(-4\)
Answer 3.
\(-25\)
Answer 4.
\(-13\)
Solution.
\begin{equation*} \begin{array}{ccc} \boldsymbol{D} \amp = \amp {4}\,\boldsymbol{B} - {2}\,\boldsymbol{C} + 2\,\boldsymbol{A}\,\boldsymbol{B} - \boldsymbol{A}\,\boldsymbol{C}\\ \amp \amp \\ \boldsymbol{D}^{\mathsf{T}} \amp = \amp {4}\cdot\boldsymbol{B}^{\mathsf{T}} - {2}\,\boldsymbol{C}^{\mathsf{T}} + 2\cdot\boldsymbol{B}^{\mathsf{T}}\,\boldsymbol{A}^{\mathsf{T}} - \boldsymbol{C}^{\mathsf{T}}\,\boldsymbol{A}^{\mathsf{T}}\\ \amp = \amp {2}\cdot\left(2\,\boldsymbol{B}^{\mathsf{T}} - \boldsymbol{C}^{\mathsf{T}}\right) + \left( 2\cdot\boldsymbol{B}^{\mathsf{T}} - \boldsymbol{C}^{\mathsf{T}}\right)\,\boldsymbol{A}^{\mathsf{T}}\\ \amp \amp\\ \amp = \amp \left(2\,\boldsymbol{B}^{\mathsf{T}} - \boldsymbol{C}^{\mathsf{T}}\right) \cdot \left({2}\,\boldsymbol{I}_{3} + \boldsymbol{A}^{\mathsf{T}}\right)\\ \amp \amp\\ \end{array} \end{equation*}
\begin{equation*} 2\,\boldsymbol{B}^{\mathsf{T}} - \boldsymbol{C}^{\mathsf{T}} = \left[ \begin{array}{rrr} {7} \amp {-3}\\ {1} \amp {12} \end{array}\right], \hspace{2cm} {2}\,\boldsymbol{I}_{3} + \boldsymbol{A}^{\mathsf{T}} = \left[ \begin{array}{rrr} {-1} \amp {-1} \\ {-2} \amp {-1} \end{array}\right] \end{equation*}
\begin{equation*} \boldsymbol{D}^{\mathsf{T}} = \left[ \begin{array}{rrr} {-1} \amp {-4}\\ {-25} \amp {-13} \end{array}\right] \end{equation*}
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