Skip to main content

Section 2.9 Powers of a matrix (negative exponents)

Suppose we have a square matrix \(A\text{.}\) For positive \(m\) and \(n\) we have proven the law of exponents: in Subsection 2.6.2.

  • \(A^n=\underbrace{A\cdots A}_{n \text{ factors}},\) (Note that \(A^1=A\))

  • \(\displaystyle A^m A^n=\underbrace{A\cdots A}_{m \text{ factors}} \underbrace{A\cdots A}_{n \text{ factors}} =\underbrace{A\cdots A}_{m+n \text{ factors}}=A^{m+n},\)

  • \(\displaystyle (A^m)^n =\underbrace{(\underbrace{A\cdots A}_{m \text{ factors}}) (\underbrace{A\cdots A}_{m \text{ factors}})\cdots (\underbrace{A\cdots A}_{m \text{ factors}})}_{n \text{ times}}=A^{mn}\)

Our goal is to extend this law for invertible matrices so that it is valid for any integer exponents. It turns out that there is only one way to do so, and we will see this in successive steps.

  • \(n=0\text{:}\) For any matrix \(A\text{,}\)

    \begin{gather*} A^mA^0=A^{m+0}=A^m\\ A^0A^m=A^{0+m}=A^m\text{,} \end{gather*}
    and so it must be that \(A^0=I\text{.}\)

  • \(m=1, n=-1\text{:}\)

    \begin{gather*} A^1 A^{-1}=A^{1+(-1)}=A^0=I\\ A^{-1} A^1=A^{-1+1}=A^0=I \end{gather*}
    so \(A^{-1}\) is the inverse of \(A\) (and hence the notation).

  • \(n=-m\text{:}\)

    \begin{gather*} A^m A^{-m}=A^{m+(-m)}=A^0=I\\ A^{-m} A^m=A^{-m+m}=A^0=I \end{gather*}
    so \(A^{-m}\) is the inverse of \(A^m.\) Notationally this says \(A^{-m}=(A^m)^{-1}\text{.}\)

The first two results have already been verified. The method for the last three is similar:

  • \((A^{-1})A=I\) implies that \(A\) is the inverse of \(A^{-1}\text{,}\) that is, \(A=(A^{-1})^{-1}.\)

  • Evaluate so the the middle factors "disappear":

    \begin{equation*} \begin{array}{rl} A^n(A^{-1})^n \amp= \underbrace{A\cdots A}_{n\text{ factors}}\ \underbrace{A^{-1}\cdots A^{-1}}_{n\text{ factors}} \\ \amp= \underbrace{A\cdots A}_{n-1\text{ factors}}(AA^{-1}) \underbrace{A^{-1}\cdots A^{-1}}_{n-1\text{ factors}} \\ \amp= \underbrace{A\cdots A}_{n-1\text{ factors}}(I) \underbrace{A^{-1}\cdots A^{-1}}_{n-1\text{ factors}}\\ \amp= \underbrace{A\cdots A}_{n-1\text{ factors}} \underbrace{A^{-1}\cdots A^{-1}}_{n-1\text{ factors}}\\ \amp=\underbrace{A\cdots A}_{n-2\text{ factors}}(AA^{-1}) \underbrace{A^{-1}\cdots A^{-1}}_{n-2\text{ factors}} \\ \amp\vdots\\ \amp= AA^{-1}\\ \amp= I \end{array} \end{equation*}

    and so \((A^{-1})^n\) is the inverse of \(A^n\text{.}\)

  • \((rA)(\tfrac 1r A^{-1}) = r\tfrac 1r A A^{-1}=1I=I\) and so \(\tfrac 1r A^{-1}\) is the inverse of \(rA\text{.}\)