Is the negative power of a matrix defined?

Question

I had a matrices exam last week and I wrote $A^{-2}$ to refer to $(A^{-1})^2$ (A being an invertible matrix). Initially, I was given the question wrong, but I told the professor that I saw in a book that $(A^{-1})^n = A^{-n} = (A^n)^{-1}$ and gave me the point.

He said that $A^{-n}$ isn't defined for matrices, which I may believe since I haven't seen it in this forum yet, but I would find having to write $(A^{-1})^n$ or $(A^n)^{-1}$ very annoying just for the fact that it isn't defined.

So, is it (defined)?

P.D.: The book is Linear Algebra by Paul Dawkins 1.

Answer 1

Yes, $A^{-n}$ is defined to be $(A^{-1})^n$ when the matrix is invertible.

For example for $$A=\begin {bmatrix}3&2\\4&5\end {bmatrix}$$

We have $$ A^{-1} = \begin {bmatrix}5/7&-2/7\\-4/7&3/7\end {bmatrix}$$

Thus $$ A^{-2} =\begin {bmatrix}5/7&-2/7\\-4/7&3/7\end {bmatrix}^2= \begin {bmatrix}33/49&-16/49\\-32/49&17/49\end {bmatrix}$$

Answer 2

You are correct that they are the same, at least for integers (rational powers of matrices are a whole other mess I'm not going to go into here).

By definition: $$(A^n)^{-1}A^n = I,$$ where $I$ is the appropriate-dimensional identity matrix. Since $A$ is square and invertible, $A^-1$ exists and it is also square. We can therefore right-multiply both sides with $(A^{-1})^n$, apply sequentially the inverses to each copy of $A$, and the result follows.

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p.s., while thinking about the problem, I also found a nice (partial) proof by induction, which I thought would be nice to include:

$\require{cancel}$ Assume you take the $n$th power of the inverse, i.e., $(A^{-1})^n$, we can show that it is the inverse of $A^n$: The base case, $n=1$, is true by definition. Now, assume that $(A^{-1})^k$ is $(A^k)^{-1}$. Then,$$(A^{-1})^{k+1} A^{k+1} = (A^{-1})^k \bcancel{A^{-1} A} A^k,$$ by definition of the matrix power. Now, by the induction hypothesis, the right-hand side is the identity. It follows that the $(A^{-1})^{k+1}$ on the left-hand side is the inverse of $A^{k+1}$, as required.