How to obtain this simplification

Question

I was studying a linear algebra problem and I encountered this step in the solution :

\begin{aligned} & \mathbf{1}^T\left(\mathbf{1 1}^T+I \sigma_v^2\right)^{-1} y =\left(\mathbf{1}^T \mathbf{1}+\sigma_v^2\right)^{-1} \mathbf{1}^T y \end{aligned}

Here $\mathbf{1}$ denote a column vector of all 1s.

Can someone clarify how this simplification is obtained. Is there a standard result in use here?

Answer 1

Since $y$ is arbitrary, the following must be true for this identity to hold: $$\mathbf{1}^T (\mathbf{1}\mathbf{1}^T+I\sigma_v^2)^{-1}=\mathbf{1}^T (\mathbf{1}^T \mathbf{1}+\sigma_v^2)^{-1}$$ Now try applying $\mathbf{1} \mathbf{1}^T+ I\sigma_v^2$ to the right on both sides of the equation.