= Solution
Insert the <eigenvalues> and <eigenvectors> into the inverse <spectral decomposition>. The first two projections contribute
$$
\frac12\mathbf e_1\mathbf e_1^{\mathsf T}+\frac14\mathbf e_2\mathbf e_2^{\mathsf T}
=\begin{pmatrix}3/8&1/8&0\\1/8&3/8&0\\0&0&0\end{pmatrix}.
$$
Adding the third projection gives
$$
\boxed{A^{-1}=\begin{pmatrix}3/8&1/8&0\\1/8&3/8&0\\0&0&1\end{pmatrix}.}
$$
Multiplication by $A$ gives the identity; in particular the upper block satisfies $\begin{pmatrix}3&-1\\-1&3\end{pmatrix}\begin{pmatrix}3&1\\1&3\end{pmatrix}=8I$.
Back to article page
