I would like to find $(I+QAQ^{T})^{-1}$ where $Q$ is the eigenvector matrix, $A$-eigenvalue matrix. I saw somewhere that it is equal to $Q(I+A)^{-1}Q^{T}$ is this so? I don't understand why should one be able to take the eigenvectors matrices out of the brackets of the inverse sign?
Inverse of $(I+QAQ^{T})$ where $Q$ is the eigenvector matrix, $A$-eigenvalue matrix
linear algebra
Best Answer
When you say that "$Q$ is the eigenvector matrix", I assume that in this context $Q$ is an orthogonal matrix (i.e. $Q^T = Q^{-1}$).
With that said, note that \begin{align} (I + QAQ^T)^{-1} & = (I + QAQ^{-1})^{-1} \\ & = (QIQ^{-1} + QAQ^{-1})^{-1} \\ & = (Q[I + A]Q^{-1})^{-1} \\ & = (Q^{-1})^{-1} (I + A)^{-1} Q^{-1} \\ & = Q(I + A)^{-1} Q^{-1} = Q(I + A)^{-1}Q^T. \end{align} The more general phenomenon is that for any "analytic function" $f(x)$ (for example, a polynomial $f(x)$, $f(x) = 1/(1 + x)$, $f(x) = e^x$), if $f(A)$ is defined "in the natural way" (via the power series of $f$ or a functional calculus), then it necessarily holds that $$ f(PAP^{-1}) = Pf(A)P^{-1} $$ for any invertible matrix $P$.