Matrix Geometric Series

For scalar geometric series, we know
$$ sum_k=0^infty x^k = dfrac11-x text and sum_k=0^infty kx^k-1 = dfrac1(1-x)^2,.$$

Does the second one extend to square matrices? We know for $A$ being a $n times n$ square matrix and $|A| < 1$, $sum_k=0^infty A^k = (I-A)^-1$. Does the following hold?

$$sum_k=0^infty k A^k-1 = (I-A)^-2 $$

2 Answers
2

Hint.
beginalign
(I-A)^-2
=&
big[(I-A)^-1big]^2
\
=&
big[sum_k=0^infty A^kbig]^2
\
=&
big[A^0+A^1+A^2+ldots +A^k_0-1+A^k_0+sum_k=k_0+1^infty A^kbig]^2
\
endalign

We know that $$sum_k=0^n x^k = fracx^n+1 - 1x-1$$

so differentiating that gives $$sum_k=1^n kx^k-1 = fracnx^n+1 - (n+1)x^n+1(x-1)^2$$

or $$(x-1)^2left(sum_k=1^n kx^k-1right) = nx^n+1 - (n+1)x^n+1$$

The evaluation map $mathbbC[x] to M_n(mathbbC) : p mapsto p(A)$ is an algebra homomorphism so we get

$$(A-I)^2left(sum_k=1^n kA^k-1right) = nA^n+1 - (n+1)A^n+I$$

The series $sum_k=1^infty kx^k-1$ converges for $|x| < 1$ so if $|A| < 1$ the series
$$sum_k=1^n k|A^k-1| = sum_k=1^n k|A|^k-1$$
also converges, and hence $sum_k=1^infty kA^k-1$ exists. On the other hand $$|nA^n+1 - (n+1)A^n| le n|A|^n+1 + (n+1)|A|^n xrightarrowntoinfty 0$$
Therefore, letting $ntoinfty$ in the above relation gives

$$(A-I)^2left(sum_k=1^infty kA^k-1right) = I$$

so $(A-I)^-2 = sum_k=1^infty kA^k-1$.

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