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Section 1.1 Matrix Transpose and Symmetric Matrices

Subsection 1.1.1 The Transpose of a matrix

Definition 1.1.1. 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 {5} \amp {-6} \amp \rceil \\ \lfloor \amp {8} \amp {7} \amp {-2} \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

\(8\)

Answer 3

\(5\)

Answer 4

\(7\)

Answer 5

\(-6\)

Answer 6

\(-2\)

Solution
\begin{equation*} \boldsymbol{A}^{\mathsf{T}} = \begin{array}{lrrr} \lceil \amp {-7} \amp {8} \amp \rceil \\ \vert \amp {5} \amp {7} \amp \vert \\ \lfloor \amp {-6} \amp {-2} \amp \rfloor \\ \end{array} \end{equation*}

Subsubsection 1.1.1.1 Properties of matrix transposes

  • Transpose of a sum.

    The transpose of a sum is equal to the sum of the tranposes.

    \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 {-1} \amp \rceil\\ \lfloor \amp {1} \amp {1} \amp \rfloor \end{array}, \hspace{0.5cm} \boldsymbol{B} = \begin{array}{lrrr} \lceil \amp {3} \amp {3} \amp \rceil\\ \lfloor \amp {1} \amp {-2} \amp \rfloor \end{array},\hspace{0.5cm} \boldsymbol{C} = \begin{array}{lrrr} \lceil \amp {-4} \amp {4} \amp \rceil\\ \lfloor \amp {-2} \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

\(-14\)

Answer 2

\(22\)

Answer 3

\(-2\)

Answer 4

\(2\)

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} {10} \amp {4}\\ {2} \amp {0} \end{array}\right], \hspace{2cm} {2}\,\boldsymbol{I}_{3} + \boldsymbol{A}^{\mathsf{T}} = \left[ \begin{array}{rrr} {-1} \amp {1} \\ {-1} \amp {3} \end{array}\right] \end{equation*}
\begin{equation*} \boldsymbol{D}^{\mathsf{T}} = \left[ \begin{array}{rrr} {-14} \amp {22}\\ {-2} \amp {2} \end{array}\right] \end{equation*}

Subsection 1.1.2 Symmetric matrices

Definition 1.1.9. Symmetric matrix.

A symmetric matrix is a square matrix that is equal to its transpose.

The following matrices are symmetric

\begin{equation*} \left[ \begin{array}{rr} 0 \amp 1\\ 1 \amp 3 \end{array}\right], \hspace{0.7cm} \left[ \begin{array}{rrr} -1 \amp 2 \amp -3 \\ 2 \amp 0 \amp 1\\ -3 \amp 1 \amp 5 \end{array}\right], \hspace{0.7cm} \left[ \begin{array}{rrrr} 1 \amp 1 \amp 4 \amp 6\\ 1 \amp -2 \amp 2 \amp 5\\ 4 \amp 2 \amp 4 \amp 3\\ 6 \amp 5 \amp 3 \amp 1 \end{array}\right] \end{equation*}

The following matrices are not symmetric

\begin{equation*} \left[ \begin{array}{rr} 1 \amp 0\\ 1 \amp 3 \end{array}\right], \hspace{0.7cm} \left[ \begin{array}{rrr} -1 \amp 2 \amp -3 \\ -2 \amp 0 \amp 1\\ -3 \amp 1 \amp 5 \end{array}\right], \hspace{0.7cm} \left[ \begin{array}{rrrr} 1 \amp 1 \amp 4 \amp 3\\ 1 \amp -2 \amp 2 \amp 7\\ 3 \amp 2 \amp 4 \amp 3\\ 6 \amp 1 \amp 3 \amp 1 \end{array}\right] \end{equation*}

Determine whether the given matrix is symmetric or not

\begin{equation*} \begin{array}{lrrrr} \lceil \amp {-3} \amp {3} \amp {3} \amp \rceil\\ \vert \amp {-3} \amp {-8} \amp {-9} \amp \vert \\ \lfloor \amp {-3} \amp {9} \amp {7} \amp \rfloor \end{array} \end{equation*}

  • Symmetric

  • Not Symmetric

\begin{equation*} \begin{array}{lrrr} \lceil \amp {3} \amp {-2} \amp \rceil\\ \vert \amp {-2} \amp {-8} \amp \vert\\ \lfloor \amp {-3} \amp {-3} \amp \rfloor \end{array} \end{equation*}

  • Symmetric

  • Not Symmetric

\begin{equation*} \begin{array}{lrrrr} \lceil \amp {3} \amp {-3} \amp {-9} \amp \rceil\\ \vert \amp {-3} \amp {-2} \amp {-8} \amp \vert\\ \lfloor \amp {-9} \amp {-8} \amp {7} \amp \rfloor \end{array} \end{equation*}

  • Symmetric

  • Not Symmetric

Answer 1

\(\text{Not Symmetric}\)

Answer 2

\(\text{Not Symmetric}\)

Answer 3

\(\text{Symmetric}\)

Subsection 1.1.3 The trace of a matrix

Definition 1.1.12. Trace.

The trace of a square matrix \(\boldsymbol{A}\text{,}\) denoted \(\text{tr}\left(\boldsymbol{A}\right)\text{,}\) is the sum of the diagonal elements of \(\boldsymbol{A}\text{.}\)

Find the \(\text{tr}\left(\boldsymbol{A}\right)\text{,}\) where

\begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrrr} 2 \amp -1 \amp 0 \amp 2\\ 3 \amp -3 \amp 4 \amp 1\\ -2 \amp 1 \amp 1 \amp 0\\ 1 \amp 2 \amp 3 \amp 4 \end{array}\right] \end{equation*}

The trace is

\begin{equation*} \begin{array}{ccc} \text{tr}\left(\boldsymbol{A}\right) \amp = \amp a_{11} + a_{22} + a_{33} + a_{44}\\ \amp = \amp 2 \hspace{0.15cm} - \hspace{0.15cm} 3 \hspace{0.15cm} + \hspace{0.15cm} 1 \hspace{0.15cm} + \hspace{0.15cm} 4 \\ \amp = \amp 4. \end{array} \end{equation*}

Find the trace of the given matrix

\begin{equation*} \begin{array}{lrrrrrr} \lceil \amp {-1} \amp {4} \amp {-1} \amp {-8} \amp {7} \amp \rceil\\ \vert \amp {7} \amp {5} \amp {-8} \amp {-8} \amp {9} \amp \vert\\ \vert \amp {-6} \amp {-6} \amp {6} \amp {-3} \amp {2} \amp \vert\\ \vert \amp {9} \amp {3} \amp {4} \amp {2} \amp {7} \amp \vert\\ \lfloor \amp {-9} \amp {0} \amp {1} \amp {1} \amp {-4} \amp \rfloor\\ \end{array} \end{equation*}

Trace =

Answer

\(8\)

Solution
\begin{equation*} \text{tr}(\boldsymbol{A}) = ({-1}) + ({5})+({6}) +({2}) + ({-4}) = {8} \end{equation*}

Subsubsection 1.1.3.1 Properties of the trace

  • Trace of a sum.

    The trace of the sum of two matrix is equal to the sum of the traces of each matrix.

    \begin{equation*} \text{tr}\left(\boldsymbol{A}+\boldsymbol{B}\right) = \text{tr}\left(\boldsymbol{A}\right) + \text{tr}\left(\boldsymbol{B}\right) \end{equation*}

    Let,

    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right], \hspace{1cm} \boldsymbol{B} = \left[ \begin{array}{rrr} 2 \amp -2 \amp 1 \\ 3 \amp -1 \amp 0 \\ 4 \amp 0 \amp 1 \end{array}\right]. \end{equation*}

    In this case,

    \begin{equation*} \begin{array}{ccc} \text{tr}\left(\boldsymbol{A}\right) \amp = \amp 1-3+1 = -1,\\ \text{tr}\left(\boldsymbol{B}\right) \amp = \amp 2-1+1 = 2.\\ \amp \amp\\ \text{tr}\left(\boldsymbol{A}\right)+ \text{tr}\left(\boldsymbol{B}\right) \amp = \amp -1 +2 = 1. \end{array} \end{equation*}

    For the sum we have,

    \begin{equation*} \boldsymbol{A}+ \boldsymbol{B} = \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right] + \left[ \begin{array}{rrr} 2 \amp -2 \amp 1 \\ 3 \amp -1 \amp 0 \\ 4 \amp 0 \amp 1 \end{array}\right] = \left[ \begin{array}{rrr} 3 \amp 0 \amp -2 \\ 4 \amp -4 \amp -2 \\ 2 \amp 0 \amp 2 \end{array}\right]. \end{equation*}
    \begin{equation*} \text{tr}\left(\boldsymbol{A}+ \boldsymbol{B}\right) = 3-4+2 = 1. \end{equation*}

    \(\,\)

  • Trace of a product.

    The trace of \(\boldsymbol{A}\,\boldsymbol{B} \) is the same as the trace of \(\boldsymbol{B}\,\boldsymbol{A}. \)

    \begin{equation*} \text{tr}\left(\boldsymbol{A}\,\boldsymbol{B}\right) = \text{tr}\left(\boldsymbol{B}\,\boldsymbol{A}\right). \end{equation*}

    Let

    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right], \hspace{1cm} \boldsymbol{B} = \left[ \begin{array}{rrr} 2 \amp -2 \amp 1 \\ 3 \amp -1 \amp 0 \\ 4 \amp 0 \amp 1 \end{array}\right]. \end{equation*}

    Then,

    \begin{equation*} \boldsymbol{A}\,\boldsymbol{B} = \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right]\, \left[ \begin{array}{rrr} 2 \amp -2 \amp 1 \\ 3 \amp -1 \amp 0 \\ 4 \amp 0 \amp 1 \end{array}\right]= \left[ \begin{array}{rrr} -4 \amp -4 \amp -2 \\ -15 \amp 1 \amp -1 \\ 0 \amp 4 \amp -1 \end{array}\right] \end{equation*}

    Giving,

    \begin{equation*} \text{tr}\left(\boldsymbol{A}\,\boldsymbol{B}\right) = -4 + 1 -1 = -4. \end{equation*}

    Similarly,

    \begin{equation*} \boldsymbol{B}\,\boldsymbol{A} = \left[ \begin{array}{rrr} 2 \amp -2 \amp 1 \\ 3 \amp -1 \amp 0 \\ 4 \amp 0 \amp 1 \end{array}\right]\, \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right] = \left[ \begin{array}{rrr} -2 \amp 10 \amp -1 \\ 2 \amp 9 \amp -7 \\ 2 \amp 8 \amp -11 \end{array}\right] \end{equation*}
    \begin{equation*} \text{tr}\left(\boldsymbol{B}\,\boldsymbol{A}\right) = -2 + 9 -11 = -4. \end{equation*}

    \(\,\)

  • Trace of a matrix-scalar product.

    The trace of the product of a matrix and a scalar is equal to the scalar times the trace of the matrix.

    \begin{equation*} \text{tr}\left(c\,\boldsymbol{A}\right) = c\,\text{tr}\left(\boldsymbol{A}\right). \end{equation*}

    Let

    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right]. \end{equation*}

    Then,

    \begin{equation*} \text{tr}\left(\boldsymbol{A}\right) = 1 -3 + 1 = -1. \end{equation*}

    If we multiply \(\boldsymbol{A}\) by \(4\text{,}\) we get,

    \begin{equation*} 4\,\boldsymbol{A} = \left[ \begin{array}{rrr} 4 \amp 8 \amp -12 \\ 4 \amp -12 \amp -8 \\ -8 \amp 0 \amp 4 \end{array}\right], \end{equation*}

    and,

    \begin{equation*} \text{tr}\left(4\boldsymbol{A}\right) = 4 -12 + 4 = -4. \end{equation*}

    Compare to

    \begin{equation*} 4\, \text{tr}\left(\boldsymbol{A}\right) = 4 \times -1 = -4. \end{equation*}

    \(\,\)

  • Trace of a transpose.

    The trace of a transpose is equal to the trace of the original matrix.

    \begin{equation*} \text{tr}\left(\boldsymbol{A}\right) = \text{tr}\left(\boldsymbol{A}^{\mathsf{T}}\right). \end{equation*}

    Let

    \begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rrr} 1 \amp 2 \amp -3 \\ 1 \amp -3 \amp -2 \\ -2 \amp 0 \amp 1 \end{array}\right]. \end{equation*}
    \begin{equation*} \text{tr}\left(\boldsymbol{A}\right) = 1 -3 + 1 = -1. \end{equation*}

    The transpose of \(\boldsymbol{A} \) is,

    \begin{equation*} \boldsymbol{A}^{\mathsf{T}} = \left[ \begin{array}{rrr} 1 \amp 1 \amp -2 \\ 2 \amp -3 \amp 0\\ -3 \amp -2 \amp 1 \end{array}\right]. \end{equation*}

    and its trace is

    \begin{equation*} \text{tr}\left(\boldsymbol{A}^{\mathsf{T}}\right) = 1 -3 + 1 = -1. \end{equation*}

    \(\, \)

Subsection 1.1.4 Homework

Link to webwork: HW 2.2