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Subsection 1.5 Powers of matrices

Definition 1.5.28. Power of a matrix.

The \(k\) power of a matrix is defined as
\begin{equation*} \boldsymbol{A}^k = \underbrace{ \boldsymbol{A} \cdot \boldsymbol{A} \cdot \boldsymbol{A} \cdots \boldsymbol{A}}_{\large{k-\text{times}}}. \end{equation*}
By definition,
\begin{equation*} \boldsymbol{A}^0 = \boldsymbol{I}, \end{equation*}
where \(\boldsymbol{I} \) denotes the identity matrix (1.1.5).

Subsubsection 1.5.1 Product of two powers

The following holds for any square matrix \(\boldsymbol{A}_{n \times n} \) and integers \(p \) and \(q\text{,}\)
\begin{equation*} \boldsymbol{A}^p \cdot \boldsymbol{A}^q = \boldsymbol{A}^{p+q}. \end{equation*}
Consider the matrix,
\begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]. \end{equation*}
  1. Calculate \(\boldsymbol{A}^2 \)
    \begin{equation*} \boldsymbol{A}^2 = \boldsymbol{A} \cdot \boldsymbol{A} = \left[ \begin{array}{cc} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] = \left[ \begin{array}{rr} 0 \amp -1 \\ 1 \amp -1 \\ \end{array} \right]. \end{equation*}
  2. Calculate \(\boldsymbol{A}^3 \)
    \begin{equation*} \begin{array}{ccl} \boldsymbol{A}^3 = \boldsymbol{A} \cdot \boldsymbol{A} \cdot \boldsymbol{A} \amp = \amp \underbrace{\left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \\ \amp = \amp \hspace{1.cm} \underbrace{\left[ \begin{array}{rr} 0 \amp -1 \\ 1 \amp -1 \\ \end{array} \right] \hspace{1.1cm} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]} \\ \amp = \amp \hspace{2.2cm} \left[ \begin{array}{rr} -1 \amp 0 \\ 0 \amp -1 \\ \end{array} \right] \end{array} \end{equation*}
  3. Calculate \(\boldsymbol{A}^5 \)
    \begin{equation*} \begin{array}{ccl} \boldsymbol{A}^5 = \boldsymbol{A} \cdot \boldsymbol{A} \cdot \boldsymbol{A}\cdot \boldsymbol{A} \cdot \boldsymbol{A} \amp =\amp \underbrace{\left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \\ \amp = \amp \hspace{1.cm} \underbrace{\left[ \begin{array}{rr} 0 \amp -1 \\ 1 \amp -1 \\ \end{array} \right] \hspace{1.1cm} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]\\ \end{array} \end{equation*}
    \begin{equation*} \begin{array}{ccl} \hspace{4.7cm} \amp = \amp \hspace{2.4cm} \underbrace{\left[ \begin{array}{rr} -1 \amp 0 \\ 0 \amp -1 \\ \end{array} \right] \hspace{1.5cm} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]\\ \amp = \amp \hspace{4.4cm} \underbrace{\left[ \begin{array}{rr} -1 \amp 1 \\ -1 \amp 0 \\ \end{array} \right]\hspace{2cm} \cdot \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]}\\ \amp = \amp \hspace{6.4cm} \left[ \begin{array}{rr} 0 \amp 1 \\ -1 \amp 1 \\ \end{array} \right] \end{array} \end{equation*}
    Alternatively, instead of multiplying \(\boldsymbol{A} \) 5-times, we can use \(\boldsymbol{A}^2 \) and \(\boldsymbol{A}^3 \) to find \(\boldsymbol{A}^5 \text{,}\)
    \begin{equation*} \begin{array}{ccl} \boldsymbol{A}^5 = \boldsymbol{A}^2 \cdot \boldsymbol{A}^3 \amp = \amp \underbrace{\left[ \begin{array}{rr} 0 \amp -1 \\ 1 \amp -1 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} -1 \amp 0 \\ 0 \amp -1 \\ \end{array} \right]}\\ \amp = \amp \hspace{1.3cm} \left[ \begin{array}{rr} 0 \amp 1 \\ -1 \amp 1 \\ \end{array} \right]. \end{array} \end{equation*}

Insight 1.5.30. Size matters - square matrix.

Consider the matrix,
\begin{equation*} \boldsymbol{A}_{3\times 2} = \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ 2 \amp 1 \\ \end{array} \right]. \end{equation*}
Find \(\boldsymbol{A}^2 \text{.}\)
This is not possible, since we cannot obtain the product,
\begin{equation*} \boldsymbol{A}_{3\times 2} \cdot \boldsymbol{A}_{3\times 2}. \end{equation*}
In general, we cannot obtain powers of a non-square matrix \(\boldsymbol{A} \text{,}\) because we cannot multiply a \(n \times m \) matrix by another \(n \times m \) matrix unless \(n = m \text{.}\)

Subsubsection 1.5.2 Power of a power

The following holds for any square matrix \(\boldsymbol{A}_{n \times n} \) and integers \(p \) and \(q\text{,}\)
\begin{equation*} \left(\boldsymbol{A}^p\right)^q = \boldsymbol{A}^{p \cdot q}. \end{equation*}
Consider the matrix,
\begin{equation*} \boldsymbol{A} = \left[ \begin{array}{rr} 1 \amp -1 \\ 1 \amp 0 \\ \end{array} \right]. \end{equation*}
Find \(\boldsymbol{A}^{10}. \)
From the previous example we know that
\begin{equation*} \boldsymbol{A}^5 = \left[ \begin{array}{rr} 0 \amp 1 \\ -1 \amp 1 \\ \end{array} \right]. \end{equation*}
We can use this and \(\boldsymbol{A}^{10} = \left(\boldsymbol{A}^5\right)^2 \) to get our solution:
\begin{equation*} \begin{array}{ccc} \boldsymbol{A}^{10} = \boldsymbol{A}^5 \cdot \boldsymbol{A}^5 \amp = \amp \left[ \begin{array}{rr} 0 \amp 1 \\ -1 \amp 1 \\ \end{array} \right] \cdot \left[ \begin{array}{rr} 0 \amp 1 \\ -1 \amp 1 \\ \end{array} \right]\\ \amp = \amp \left[ \begin{array}{rr} -1 \amp 1 \\ -1 \amp 0 \\ \end{array} \right]. \end{array} \end{equation*}
Find \(\boldsymbol{A}^6\) where
\begin{equation*} \boldsymbol{A} = \left[ \begin{array}{cc} {3} \amp 0 \\ {-1} \amp {1} \end{array}\right]. \end{equation*}
\(\,\) \(\lceil\) \(\rceil\)
\(\boldsymbol{A}^6 =\) \(\lfloor\) \(\rfloor\)
\(\,\) \(\,\) \(\,\) \(\,\)
Answer 1.
\(729\)
Answer 2.
\(0\)
Answer 3.
\(-364\)
Answer 4.
\(1\)
Solution.
\begin{equation*} \boldsymbol{A}^6 = \boldsymbol{A}^3 \cdot \boldsymbol{A}^3 =\left[ \begin{array}{cc} {27} \amp {0} \\ {-13} \amp {1} \end{array}\right] \cdot \left[ \begin{array}{cc} {27} \amp {0} \\ {-13} \amp {1} \end{array}\right] = \left[ \begin{array}{ccc} {729} \amp {0} \\ {-364} \amp {1} \end{array}\right]. \end{equation*}
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