Proving that $ (A + I)^n=(2^n-1)A + I $

Question

Assume that $ A^2 = A$ , where $A$ is a square matrix.

Then prove $(A + I)^n=(2^n-1)A + I $

Could someone help ?

Answer 1

Note that for all $n\ge 1$ we have $A^n = A$ and hence $$(A + I)^n= \sum_{j=0}^{n}{n\choose j}A^j=I+A\sum_{j=1}^{n}{n\choose j}1^j=(2^n-1)A + I$$

Or By induction for $n=1$ the result follows.

Assume $$(A + I)^n=(2^n-1)A + I$$ then $$(A + I)^{n+1} = (A + I)^n (A + I) = ((2^n-1)A + I)(A + I) \\=(2^n-1)A + I +(2^n-1)A^2 + A \\= (2^n-1)A + I +(2^n-1)A + A \\= (2^n-1)A + I +2^nA \\=(2^{n+1}-1)A + I $$

Answer 2

By indution if $(A+I)^n=(2^n-1)A+I$ and $A^2=A$, we have \begin{eqnarray} (A+I)^{n+1}&=&(A+I)^n(A+I)=((2^n-1)A+I)(A+I)\\ &=&(2^n-1)A^2+A+(2^n-1)A+I\\ &=&2(2^n-1)A+A+I\\ &=&(2^{n+1}-2)A+A+I\\ &=&(2^{n+1}-1)A+I \end{eqnarray}

Answer 3

$A$ and $I$ commute, hence the binomial theorem (for matrices) gives $$(A+I)^n = \sum_{k=0}^n{n \choose k} A^{n-k}I^k$$

$$= I + \sum_{k=0}^{n-1}{n \choose k}A \tag{1}$$

$$= I + (2^n - 1)A \tag{2}$$

$(1)$

see comment

$(2)$

$2^n = (1+1)^n = \sum_{k=0}^n {n \choose k}$