On Neumann-series of matrices
Let $A \in \mathbb{R}^{n \times n}$ and we denote with $\rho(A)$ the spectral radius of $A$ and with $I_n \in \mathbb{R}^{n \times n}$ the identitiy matrix.
Applying Carl Neumann's result on matrices we know that if $\rho(A)<1$ then $(I-A)^{-1}$ matrix exists and $$(I_n-A)^{-1}=\sum_{k=0}^{\infty} A^k.$$
Is it true backwards? So if $(I-A)^{-1}$ matrix exists and $(I_n-A)^{-1}=\sum_{k=0}^{\infty} A^k$, then
$$\rho(A)<1?$$
If it is how could we prove it? If not: give a counter-example with a positive $A$ matrix if it's possible.
$\endgroup$1 Answer
$\begingroup$It is true backwards. One can prove by contradiction.
If $(I_n−A)^{−1}=\sum\limits_{k=0}^\infty A^k$ exists and if $\lambda_{max} > 1$ consider $$Q^{-1}(I_n−A)^{−1} Q = \sum\limits_{k=0}^\infty \Lambda^k $$ where $Q$ is an invertible matrix that diagonalizes $A$. The first diagonal element of the matrix on the LHS can be written as
$$e_1^T Q^{-1}(I_n−A)^{−1} Q e_1 = \sum\limits_{k=0}^\infty e_1^T\Lambda^k e_1 = \sum\limits_{k=0}^\infty \lambda_{max}^k = \infty,$$
where I have assumed without loss of generality that the first diagonal element of $\Lambda$ is $\lambda_{max}$. The fact that one of the values of the matrix is $\infty$ violates our assumption that $(I_n−A)^{−1}$ exists.
edit : for non-diagonalizable matrices, we use the Jordan normal form. $$P^{-1}(I_n−A)^{−1} P = \sum\limits_{k=0}^\infty J^k, $$ where $J$ is an upper triangular matrix containing the eigenvalues on the diagonal. Since the product of two upper triangular matrices are upper triangular and the diagonal of the product matrix is the product of the diagonal elements of the individual matrices, we can again write
$$e_1^T P^{-1}(I_n−A)^{−1} P e_1 = \sum\limits_{k=0}^\infty e_1^T J^k e_1 = \sum\limits_{k=0}^\infty \lambda_{max}^k = \infty,$$
which is a contradiction of the original statement
$\endgroup$ 4
