Math Equations

Monday, October 27, 2014

Elegant matrix inversion

I had an exam today. One question inspired me. Let \(\mathbf{E}_n\hspace{-3pt}: n\times n\) and \(\mathbf{e}_{ij}=1\). Then the inverse of \(\mathbf{A}_n = \mathbf{E}_n - \mathbf{I}_n\) is \(\mathbf{A}_n^{-1} = \frac1{n-1}\mathbf{E}_n-\mathbf{I}_n\) if \(2\leq n\). There are two interesting things about this: The inverse can be expressed using matrix operations, and the matrix operations can be used to verify that \(\mathbf{A}^{-1}\) is actually an inverse.

To verify the result, note first that \(\mathbf{E}_n^2=n\mathbf{E}_n\). This identity can be seen as the workhorse of the method. Check that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}\).
\[\begin{align}\mathbf{A}_n^\phantom{-1}\hspace{-4pt}\mathbf{A}_n^{-1} &= (\mathbf{E}_n - \mathbf{I}_n)\left(\frac1{n-1}\mathbf{E}_n-\mathbf{I}_n\right) \\&= \frac n{n-1}\mathbf{E}_n-\mathbf{E}_n-\frac1{n-1}\mathbf{E}_n+\mathbf{I}_n = \mathbf{I}_n\end{align}\]

I'll have to think about this some more sometime. Can other simple matrices be inverted and expressed using matrix operations? What are some useful workhorses? Why has no one told me about this before?