Geometric sum of matrices I have a markov matrix $$A=\begin{bmatrix}
\lambda_1 & 1-\lambda_1\\
1-\lambda_2 & \lambda_2
\end{bmatrix}$$and I want to calculate $I+A+A^2+A^3+...$I read somewhere that if $|A|<1$ then the sum is $(I-A)^{-1}.$ The condition holds in my case since both of my $\lambda$ are probability. However, the matrix $I-A$ is not invertible. I am very confused about this.
 A: There's an extra condition. The dedication of the formula you cite also requires that $I-A$ be nonsingular. You have to evaluate this sum in another way. I think you can use Jordan-Chevalley decomposition.
A: The norm being used here is the Frobenius norm : If $|A|_F < 1$ then the series converges. However, for a matrix of the given form, we have :
\begin{align}
|A|_F^2 &= (1-\lambda_1)^2 + \lambda_1^2 + \lambda_2^2 + (1-\lambda_2)^2 \\
&\geq 2^{-1}(1 - \lambda_1 + \lambda_1 + \lambda_2 + 1-\lambda_2) \tag{RMS-AM}
\\&= 1  
\end{align}
Therefore it is not at all guaranteed that the series corresponding to $(I-A)^{-1}$ must converge! Indeed, it is even easier to see that $(I-A)$ is a rank one matrix for the given $A$ so its inverse will not exist.
Also note that even if your norm was different, all the norms on matrices are equivalent, so taking a different still won't take $1+A+A^2 + \cdot $ to $(I-A)^{-1}$ as a convergent sum.

Indeed, the answer to the question then is that $I+A+A^2 + ...$ is not a convergent series (i.e. the result is a matrix with possibly infinite entries). The reason is as follows : suppose it converged to a matrix $B$, then it follows that $|A^n|$ must go to zero in some (hence every) norm, since the tail of a convergent series converges in norm to zero.
However, for the given matrix, it is easy to see that for all $n$, the matrix $A^n$ has all its row sums equal to $1$, so it has an eigenvalue $1$ with eigenvector $[1,1]$. Thus, in the usual matrix norm, the series $A^n$ doesn't go to zero! So there's no way the series is convergent.
