- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.21 Miscellaneous review exercises on matrices
5. $\quad$ If $A^2 = A,$ prove that $(A + I)^k = I + (2^k - 1)A.$
$\quad$ Proof. $\quad$ First, we will prove by induction that for any integer $k \geq 1,$ $A^k = A.$ When $k = 1, A^1 = A.$ When $k = 2,$ we have $A^2 = A$ by our given information. Now, assume $A^n = A$ for some $n \geq 2,$ then \begin{align*} A^{n + 1} &= A^nA = AA = A^2 = A. \end{align*} And since this is true for $n + 1,$ by induction it holds for any integer $k \geq 1.$
$\quad$ Now, if we recall from Volume 1, Section I-4.10, Exercise 4, the binomial theorem for scalars $a, b$ is given by \begin{align*} (a + b)^n &= \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}. \end{align*} In the special case where $a = b = 1,$ we have \begin{align*} 2^n &= \sum_{k=0}^n \binom{n}{k}. \end{align*} If we subtract the first term of the sum from both sides, we have \begin{align*} 2^n - 1 &= \sum_{k=1}^n \binom{n}{k}. \end{align*} Then, applying the distributive laws of matrix multiplication, we can extend the binomial theorem to the matrix product $(A + I)^k$ as follows: \begin{align*} (A + I)^k &= \sum_{p=0}^k \binom{k}{p} A^p I \\ &= I + \sum_{p=1}^k \binom{k}{p} A \\ &= I + (2^k - 1) A. \quad \blacksquare \end{align*}