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?
2026-03-25 03:01:38.1774407698
How to calculate square matrix to power n with induction
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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 $n \in \Bbb N$.