Let A be a square complex matrix. Show that I-A^(m+1) = (I-A)(I+A+...+A^m).

Question

Let A be a square complex matrix. Show that $$I-A^{m+1} = (I-A)(I+A+...+A^m)$$

My attempt is as follows:

Proof by induction. Let $m=0$. Then $(I-A)=(I-A)$. Assume true for any $m$. Then consider

$$I-A^{(m+1)+1}=I-A^{m+1}A$$

I want to apply what we have for $I-A^{m+1}$ but I can't just separate this from the other $A$ can I?

Answer 1

Hint: $\;I\color{red}{-A^m+A^m}-A^{m+1} = \underbrace{(I-A)(I+A+\ldots+A^{m-1})}_{\text{by the induction hypothesis}} + (I-A)A^{m}\,$.

Answer 2

You can deduce this from the telescopic sum also. Consider $$A^{k+1}-A^k= (A-I) A^k $$

you can apply the sum $\sum\limits^{n}_{k=1}$ in both sides. The first is telescopic (many terms cancel ), then you find

$$A^{n+1}-I=(A-I)\sum\limits^{n}_{k=1} A^k $$