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I know this is a very common question to ask when one is making their way into Linear Algebra (i.e. given a matrix, find the result of that matrix to the nth-power).

I'm given this matrix: $$ A= \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right]$$

I'm asked to compute, by hand, the matrix $A^{144}$

What I tried:

I calculated $A^2$, $A^3$, $A^4$ and $A^5$ and tried to find a pattern, so that I could first find the more general $A^n$ expression, and then make $n=144$. This is what I got:

$$ A^2= \left[ {\begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array} } \right]$$ $$ A^3= \left[ {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right]$$ $$ A^4= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$$ $$ A^5= \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right]$$

So there seems to be kind of a "circular pattern", where $A^6=A^2$ and therefore, since $144/6=24$, I'm suspecting that: $$¿\,\boxed{A^{144}=A^2} \,?$$

However, I was hoping to find the more general $A^n$ matrix first and confirm the above result.

But, I can't seem to find a function such that: $$f(n) = \left\{ \begin{array}{ll} 0 & \mbox{if n is odd} \\ 1 & \mbox{if n is even} \end{array} \right. $$ This function would allow me to sort out the (1,1) and (2,2) elements of $A^n$ being 0 when n is odd (and the (1,2) and (2,1) being 0 when n is even).

I'm a bit confused at this point, your help is greatly appreciated.

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  • $\begingroup$ Try $n+1\pmod{2}$. $\endgroup$ – Steve D Sep 28 '17 at 20:05
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    $\begingroup$ That matrix turns out to be the representation of $i$ (i.e. it represents counterclockwise multiplication by $\pi/2$), so that $A^4=I$. You can work this out by complex-number diagonalization if you want. $\endgroup$ – Ian Sep 28 '17 at 20:06
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    $\begingroup$ Note that $A^4 = I$ $\endgroup$ – Arthur Sep 28 '17 at 20:06
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    $\begingroup$ @Ian What do you mean by "the representation of i"? I'm a bit of a newcomer to this world. I appreciate your help. $\endgroup$ – Jose Lopez Garcia Sep 28 '17 at 20:08
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    $\begingroup$ That’s right. Use Ian’s hint to figure out which ones to use. De Moivre’s formula might be helpful. $\endgroup$ – amd Sep 28 '17 at 20:41
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You have a weird problem, you say you can't find a function such that it equals $0$ on odd integers and $1$ on even ones, yet saying this is defining such a function.

Otherwise, you saw that $A^4 = I$ (the identity matrix), so just make a division :

$144 = 4\times36$ so that $A^{144} = (A^4)^{36} = I^{36} = I$

A nice interpretation in $\mathbb{R}^2$ is that your matrix represents a rotation by $\frac{\pi}{2}$ around the origin in the canonical basis. Then, rotating $144$ times results in not rotating at all...

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As you wrote in you post: $A^4 = I$. Therefore: $A^{4k} = I$ for all $k \in \mathbb{N}$.

Because $144 = 4 \times 36$, you have :

$$A^{144} = I.$$

From the results in your post, you can deduce that:

$$ \forall k \in \mathbb{N}, \; A^{4k} = I, \; A^{4k+1} = A, \; A^{4k+2} = - I \; \text{and} \; A^{4k+3} = -A. $$

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Your analysis shows that $A^n = A^{n \text{ mod } 4}$. Since you know $A^0,A^1, A^2, A^3$ you then have a general formula. Note that $144 \text{ mod } 4 = 0$.

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Note that $A^4 = I$, and $A^{144} = (A^4)^{36}$, and you're done. If you really want to go with that $A^6 = A^2$ instead, then we have that $$A^{144} = (A^6)^{24} \\ = (A^2)^{24} = A^{48} \\ = (A^6)^8 = (A^2)^8 \\ = A^{16}= A^4\times A^{12}\\ = A^4 \times(A^6)^2 = A^4\times(A^2)^2\\ = A^8 = A^2\times A^6\\ = A^2\times A^2 = A^4$$ This seems a much more roundabout way to get the same result.

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Even when there isn't a function, you can just define and use it as you just did. But regardless, your answer can be found in modular arithmetic.

$f(n) \equiv n+1 \pmod{2}$

Here, this represents the remainder when divided by 2. If you need further help, let me know.

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  • $\begingroup$ That's technically not what $\equiv$ means. (I know, it's an annoying discrepancy between how mathematicians talk about modular arithmetic and how programmers use it, but still, that's not what it means.) $\endgroup$ – Ian Sep 28 '17 at 20:12
  • $\begingroup$ @Ian It was the notation I was taught for modular arithmetic. Though I've also seen it used for function identities. $\endgroup$ – Mad Maniac Sep 28 '17 at 20:13
  • $\begingroup$ @Ian Okay, I see your point. Assuming you mean it outputs a class of numbers as opposed to a single nunber. $\endgroup$ – Mad Maniac Sep 28 '17 at 20:19
  • $\begingroup$ In math notation it doesn't "output" anything. It's just a relation. $\endgroup$ – Ian Sep 28 '17 at 20:29

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