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This question relates to an answer of mine, I suspect I got it slightly wrong and that it should be:

$$E={\bf s}^T\left(\sum_{k=1}^{\infty}(k+1)({\bf PD})^k\right){\bf P s}$$

What would be really curious if that was the case is that it would be the termwise derivative of the power series:

$${\bf s}^T\left(\sum_{k=1}^{\infty}({\bf PD})^{k+1}\right){\bf Ps}$$

Which we recognize from the formula for the geometric series: $$\frac{1}{1-x} = 1+x+x^2+\cdots$$ It's derivative being $$\frac{\partial}{\partial x}\left\{\frac{1}{1-x} \right\} =\frac{1}{(1-x)^2}$$

Now with this logic, would it make any sense to interpret the above expression as $${\bf s}^T(({\bf I-(PD)})^2)^{-1}{\bf Ps}$$

or am I just stumbling about, muttering non-sense?

Community
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mathreadler
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1 Answers1

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Let $F(M)=\sum_{k=0}^\infty(k+1)M^k$ and $G(M)=((I-M)^2)^{-1}$.

Outline of steps to prove that $F(M)=G(M)$ for all square matrices $M$ for which the series in $F$ converges:

  1. Prove that $F(D)=G(D)$ whenever $D$ is a diagonal matrix.

  2. Then show that $F(M)=G(M)$ whenever $M$ is diagonalizable.

  3. Then extend that to the non-diagonalizable matrices by using the fact that the diagonalizable matrices are dense in the set of all complex matrices and that $F-G$ is continuous.

wlad
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