- 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
2.16 Exercises
6. $\quad$ Let $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.$ Verify that $A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and compute $A^n.$
Solution. $\quad$ First, we compute $A^2 = AA:$ \begin{align*} A^2 &= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \end{align*} Now, by definition of integral powers for a square matrix $A,$ we know that $A^n = AA^{n-1},$ with $A^0 = I.$ Thus we have proven for $0 \leq n \leq 2$ that $A^n$ is given by the matrix: \begin{align*} A^n &= \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \end{align*} Now, assume that this is true for some $n \geq 2.$ We will show that it is also true for $n + 1.$ In the case of $n = k,$ we have \begin{align*} A^k &= \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \end{align*} Then, by integral powers of $A,$ we have $A^{k+1} = AA^k,$ or: \begin{align*} A^{k+1} &= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 + 0 & k + 1 \\ 0 + 0 & 0 + 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & k + 1 \\ 0 & 1 \end{bmatrix} \end{align*} Then, by induction, if this is true for $k + 1,$ it is true for any integer $n \geq 0. \quad \blacksquare$