3
$\begingroup$

I'm trying to figure out why the calculation below works. I do know that $(A^T)^{-1} = (A^{-1})^T$.

The matrix A = $\begin{pmatrix} 1 & -1 & 0 \\ 1 & 1 & -1\\ 1 & 2 & -1 \end{pmatrix} $ has the inverse $\begin{pmatrix} 1 & -1 & 1 \\ 0 & -1 & 1\\ 1 & -3 & 2 \end{pmatrix} $

Another matrix B = $\begin{pmatrix} 1 & 1 & 1 & 0 \\ -1 & 1 & 2 & 0 \\ 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & 4 \end{pmatrix} $ which has the transpose of matrix A embedded within it has the inverse $\begin{pmatrix} 1 & 0 & 1 & 0 \\ -1 & -1 & -3 & 0 \\ 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1/4 \end{pmatrix} $.

Clearly, the inverse of B can be worked out quicker by using the result of the inverse of A and then changing 4 to its inverse (1/4). Does anyone know the actual rules that have been applied here? Is this a special case because the newly added row 4 and column 4 all have zeroes apart from 4? I just want to know why this works and in what context can I apply this trick?

Thanks

$\endgroup$
6
$\begingroup$

This is a simple consequence of the following:

If $B \in \mathbb R^{(m+n)\times(m+n)}$ is of the form $$B = \pmatrix{B_{11}&0\\0&B_{22}}$$ with $B_{11}\in\mathbb R^{m\times m}$ and $B_{22}\in\mathbb R^{n\times n}$ both invertible, then $B$ is invertible with $$B^{-1} = \pmatrix{B_{11}^{-1}&0\\0&B_{22}^{-1}}$$

To prove it, note that $$\pmatrix{A_{11}&0\\0&A_{22}}\pmatrix{B_{11}&0\\0&B_{22}}=\pmatrix{A_{11}B_{11}&0\\0&A_{22}B_{22}}$$

Iterating the theorem gives an analogous result for block-diagonal matrices with more than two blocks.

$\endgroup$
  • 1
    $\begingroup$ @John Always happy to help - especially when I see that the OP has put some thought into the question ;) $\endgroup$ – AlexR May 26 '15 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.