# How to calculate square matrix to power n with induction [duplicate]

How could i possibly calculate square matrix to power n with induction for lets say this square matrix $A =\begin{bmatrix} 1 & 2\\ 0& 1 \end{bmatrix}$I cant use Cayley-Hamilton theorem and i need to use induction.Any ideas on what i could do?

• Do the first few powers, see the pattern, and conjecture a formula for what you see. Then prove that formula using induction. Commented Mar 12, 2018 at 18:38
• ohhh i was actually doing that but then i thought there would be something more general.Thanks a lot! Commented Mar 12, 2018 at 18:51

Note that

$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = I + N, \tag1$

where

$N = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix} = 2 \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag 2$

let

$Z = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag 3$

then from (1)-(3),

$A = I + 2Z; \tag{4}$

we further note that

$Z^2 = N^2 = 4Z^2 = 0 \tag{5}$

by an easy calculation. Since $IZ = Z = ZI$ we have

$A^2 = (I + 2Z)^2 = I^2 + 2(2Z) + 4Z^2 = I + 4Z; \tag 6$

also,

$A^3 = (I + 2Z)^3 = (I + 2Z)(I + 2Z)^2$ $= (I + 2Z)(I + 4Z) = I^2 + 6Z + 8Z^2 = I + 6Z; \tag 7$

based upon (4), (6), (7) we make the inductive hypothesis

$A^k = I + 2kZ; \tag 8$

then

$A^{k + 1} = AA^k = (I + 2Z)(I + 2kZ) = I^2 + 2kIZ + 2ZI + 4kZ^2 = I + 2(k + 1)Z; \tag 9$

we thus conclude that

$A^n = I + 2nZ = \begin{bmatrix} 1 & 2n \\ 0 & 1 \end{bmatrix} \tag{10}$

for all $n \in \Bbb N$.