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Subsection 2.3 The trace of a matrix

Definition 2.3.46. 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 {2} \amp {-6} \amp {1} \amp {4} \amp \rceil\\ \vert \amp {-3} \amp {1} \amp {5} \amp {6} \amp {-8} \amp \vert\\ \vert \amp {6} \amp {9} \amp {6} \amp {5} \amp {-9} \amp \vert\\ \vert \amp {8} \amp {-9} \amp {3} \amp {4} \amp {1} \amp \vert\\ \lfloor \amp {-4} \amp {8} \amp {-2} \amp {-5} \amp {-2} \amp \rfloor\\ \end{array} \end{equation*}
Trace =
Answer.
\(8\)
Solution.
\begin{equation*} \text{tr}(\boldsymbol{A}) = ({-1}) + ({1})+({6}) +({4}) + ({-2}) = {8} \end{equation*}

Subsubsection 2.3.1 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*}
    \(\, \)
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