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!

  • 4
    $\begingroup$ In addition to powers of elements in a group being well-defined, every power of a given element in a group commutes with every other power of the same element; the subgroup generated by a single element is cyclic. $\endgroup$
    – anon
    Dec 19, 2012 at 2:30
  • 2
    $\begingroup$ Remember that multiplication is associative, so $A^n*A^k = A^{n+k} = A^{k+n} = A^k *A^n$! $\endgroup$ Dec 19, 2012 at 2:43

3 Answers 3


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.

  • 3
    $\begingroup$ How simple and clear. I think because of this post I now have a greater insight to the associativity rules which before I took for granted. Thank you. $\endgroup$ Dec 19, 2012 at 2:34
  • 4
    $\begingroup$ Your intuitions about associativity should come from your intuitions about performing a sequence of actions in some order (matrices perform geometric actions like rotation and reflection but the intuition is more general). For example, the sequence of actions ((put on socks, put on shoes), go out the door) is the same as the sequence of actions (put on socks, (put on shoes, go out the door)). From this perspective the fact that powers are well-defined means that repeating an action $n$ times is a well-defined thing to do. $\endgroup$ Dec 19, 2012 at 3:05
  • 4
    $\begingroup$ Another way of restating my point is that you shouldn't think of multiplication as a binary operation. Multiplication is a family of $n$-ary operations: the $n$-ary one takes as input an ordered list of $n$ matrices (or actions, or whatever) and returns as output a matrix, and this family of operations satisfies compatibility conditions which assert that the meaning of multiplication is invariant under splitting up an ordered list into a list of lists. All of this data happens to be determined by the binary operation, but the binary operation is only convenient, not fundamental. $\endgroup$ Dec 19, 2012 at 6:23
  • 1
    $\begingroup$ BTW, Qiaochu: Welcome back! $\endgroup$
    – amWhy
    Dec 20, 2012 at 2:05

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!

  • 1
    $\begingroup$ Nice tip on reducing the calculation after finding a $-I$, thanks! $\endgroup$ Dec 19, 2012 at 2:37

I want to give a different solution to this. Note that there is an isomorphism between $\mathbb C$ and the group of all matrices of the form $\begin{pmatrix} a & -b \\b & a\\\end{pmatrix}$, where $a^2+b^2\neq 0$. The bijection is $a+ib\mapsto \begin{pmatrix} a & -b \\b & a\\\end{pmatrix}$. Now the given matrix corresponds to $i\in \mathbb C$ whose order is $4$. Hence we are done.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .