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I am working on a problem involving basic abstract algebra/group theory and am getting confused. I am following an online course by Dr. Bob found here, and am currently on assignment two.

My difficulty lies with problem 1b where I am given a matrix $A=$ $\left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)$ and asked to find its order.

Now I am fairly sure that matrix multiplication is not commutative so this makes me suspect that there are either multiple answers or a convention we must adopt (which I dont think he mentioned). If I multiply on the right I get $A\cdot A = -I, A^3 = A\cdot A^2 = \left( \begin{array}{ccc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$, and $A^4 = A\cdot A^3 = I$ so $|A| =4$.

Now when I do this on the by multiplying on the left by $A$ I get the same answer, but my intuition says this is a coincidence because of the trivial chosen matrix.

Is it true in general that the order of elements in $GL(2,\mathbb{R})$ is the same regardless of which side you multiply on, or are there criterion when this property holds? Finally, since I'm guessing that this is just a special case situation, which side do I multiply on when asked to find the order of an element?

Thanks for the help!

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    Remember that multiplication is associative, so $A^n*A^k = A^{n+k} = A^{k+n} = A^k *A^n$!2012-12-19

2 Answers 2

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Matrix multiplication is not commutative, but it is associative, so taking powers make sense: $A^3 = A \cdot (A \cdot A) = (A \cdot A) \cdot A$, and so forth. More generally, $A^n$ is well-defined.

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    BTW, Qiaochu: Welcome back!2012-12-20
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A square matrix commutes with itself, with respect to both matrix addition and multiplication, and we have associativity to justify this:

By associativity, we have, e.g. $A^4 = A\cdot A^3 = A\cdot (A \cdot A^2) = A\cdot(A\cdot(A\cdot A)) = A\cdot ((A\cdot A)\cdot A) $ $= (A \cdot (A \cdot A))\cdot A = ((A\cdot A)\cdot A)\cdot A = (A^2\cdot A) \cdot A = A^3\cdot A = A^4$

In general: $A^n = A\cdot A^{n-1} = A^2 \cdot A^{n-2} = \cdots = A^{n-2}\cdot A^2 = A^{n-1} \cdot A$...

Indeed, note that in your present exercise, after having computed $A^2 = -I$, we have that $A^4 = A^2\cdot A^2 = -I \cdot -I = (-1)^2 I^2 = I$ which could have saved you a lot of work!

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    Nice tip on reducing the calculation after finding a $-I$, thanks!2012-12-19