0

Why does the invert of matrix $I-\gamma A$ always exist? Where $I$ is the identity matrix, $\gamma > 1$ a positive scalar, and A is an n by n right stochastic matrix (a real square matrix, with each row summing to one).

erfaun
  • 159

1 Answers1

1

This isn't true. In fact, for every $\gamma\ge1$, $I-\gamma A$ is singular when $A=\pmatrix{1&0\\ 1-\frac1\gamma&\frac1\gamma}$.

However, it is true that $I-\gamma A$ is invertible when $|\gamma|<1$. This is because the spectral radius of every stochastic matrix is equal to $1$, so that $|\gamma\lambda(A)|<1$ and $$ |\lambda(I-\gamma A)| =|1-\gamma\lambda(A)| \ge\left|1-|\gamma\lambda(A)|\right|>0. $$

Alternatively, since $\rho(\gamma A)=|\gamma|\rho(A)<1$, there exists some submultiplicative matrix norm such that $\|\gamma A\|<1$. Therefore the Neumann series $S=I+\gamma A+(\gamma A)^2+(\gamma A)^3+\cdots$ converges. Then $(I-\gamma A)S=I$ and hence $I-\gamma A$ is invertible.

user1551
  • 149,263