Power series of the inverse of a sum of matrices

Question

Is there exist a power series expansion for the inverse of a matrix that has the form $(\mathbf{K}^T\mathbf{K}+\mathbf{A})^{−1}$?

where, $\mathbf{K}_{m \times n}$; $\mathbf{A}_{n \times n}$ is not invertible. However, the sum, $\mathbf{K}^T\mathbf{K}+\mathbf{A}$ is.

Thanks!

Answer 1

If $K^⊤K$ is invertible and the spectral radius $ρ\big((K^⊤K)^{-1}A\big) < 1$, then

$$ \big(K^⊤K + A\big)^{-1} = \big((K^⊤K)( + (K^⊤K)^{-1}A)\big)^{-1} = \Big(∑_{k=0}^{∞} \big((K^⊤K)^{-1}A\big)^k \Big)(K^⊤K)^{-1}$$

Answer 2

Let $B=K^TK+A$. By the Cayley-Hamilton theorem, a square matrix satisfies its own charactertstic equation, so $$B^n+\beta_1B^{n-1}+...+\beta_{n-1}B+\beta_nI=0$$ where $$\beta_i=(-1)^i\text{ sum of all the principal minors of $B$ of order $i$}, 1 \le i \le n.$$. In particular, $\beta_n=(-1)^n\det(B)$, so, since $B$ is invertible, $\beta_n \ne 0.$ $$B(B^{n-1}+\beta_1B^{n-2}+...+\beta_{n-1}I)=-\beta_nI$$ Thus,$$B^{-1}=-\frac{1}{\beta_n}(B^{n-1}+\beta_1B^{n-2}+...+\beta_{n-1}I)$$There is a nice recursive formula, due to Bocher, for obtaining the $\beta_i$ in terms of the trace of the powers of $B$. For efficient calculation you can evaluate the polynomial in $B$ by what is essentially synthetic division.