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Question. To put the question in another way, let $G$ be the set of all invertible linear maps of $\mathbb{R}^2\to\mathbb{R}^2$, and it is not hard to show that $G$ is a group under composition, and the question asks for those elements of $G$ which have order $4$.

My Attempt. First of all, I can show that every element of $G$ is equivalent to left-multiply by a $2 \times 2$ invertible matrix. Then, I was able to find elements of $G$ of order $2$ by brutal-force algebra. However, the brutal-force method becomes tedious to be used to tackle the given question. I know two elements of $G$ which have order $4$, left-multiplying by a matrix which represents either a clockwise rotation of $90$ degrees or an anticlockwise rotation of $90$ degrees, but are they the only elements of order $4$? I have no clue how to proceed.

Note. I am merely asking for a hint rather than a solution. Thank you!

IncredibleSimon
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  • Do you know the minimal polynomial of a matrix? Also the characteristic polynomial is very useful here. – Joshua Tilley Feb 08 '24 at 09:52
  • To think about it another way: can you figure out the determinant and the trace of such a matrix? – Joshua Tilley Feb 08 '24 at 09:53
  • Hi Joshua. I have no idea what the minimal polynomial of a matrix is, but I do know the characteristic polynomial which is used to find the eigenvalues of a matrix. I'm a first year second term undergraduate in maths. @JoshuaTilley – IncredibleSimon Feb 08 '24 at 09:55
  • Do you know the Cayley Hamilton theorem? This is very useful here. – Joshua Tilley Feb 08 '24 at 09:57
  • Well, let such matrix be $A$, since $A^4=I$, its determinant must be either $1$ or $-1$. For the trace, I have no idea for now; I need some time to think about it. @JoshuaTilley – IncredibleSimon Feb 08 '24 at 09:58
  • Yes. I do know the Cayley Hamilton theorem which states that a matrix "satisfies" its characteristic polynomial. Okay, I see. I'm trying now. @JoshuaTilley – IncredibleSimon Feb 08 '24 at 10:01
  • Hi Joshua. Do you mind providing a further hint? When I was working with characteristic polynomials and the Cayley Hamilton theorem, it seemed to boil down to equations of entries of the wanted matrix, which is essentially similar to the brutal-force approach. Thanks. @JoshuaTilley – IncredibleSimon Feb 08 '24 at 11:08
  • I've posted an answer to explain. You should not need to use any calculations with entries. – Joshua Tilley Feb 08 '24 at 12:10

3 Answers3

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The Cayley Hamilton theorem says that a matrix $A$ satisfies its own characteristic polynomial. In the case of a $2\times 2$ matrix $A$, this is

$$\chi_A(t)=t^2-\operatorname{tr}(A) t+\det(A).$$

You are looking for a matrix $A$ such that $A^4=\operatorname{id}$ and by Cayley Hamilton, you know $A^2-\operatorname{tr}(A) A+\det(A)\operatorname{id}=0$. Also, as you noted, $\det(A)=\pm 1$. Try to get an equation of the form $\alpha A+\beta\operatorname{id}=0$ from the two equations

\begin{align*} A^4-\operatorname{id}&=0\\ A^2-\operatorname{tr}(A)A+\det(A)\operatorname{id}&=0 \end{align*}

Joshua Tilley
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  • Thank you, Joshua! With your guide, I was able to obtain that if $A$ is an element of $G$ of order $4$, then $A^2=-I$. Then, I resorted to specific entries to solve for $A$. – IncredibleSimon Feb 08 '24 at 14:18
  • @IncredibleSimon indeed this answer is neater. given that you are doing a course, do you know how to prove that $A^2=-I$? feel free to reply if you want/need further clarity on this. – Chris Sanders Feb 08 '24 at 16:15
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Hint as requested:

Any complex matrix is triangularisable i.e. is similar to either $\begin{pmatrix} a & 0\\ c & a \end{pmatrix}$ or $\begin{pmatrix} a & 0\\ 0 & d \end{pmatrix}$ where $a\neq d$.

Now prove that in the case of $\begin{pmatrix} a & 0\\ c & a \end{pmatrix}$, if $a\neq0$ and $c\neq0$, then any power of that triangular matrix will never be diagonal (except the zero$^{th}$ power of course).

The rest is easy if you know the above.

Chris Sanders
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You are asking for all $2×2$ real matrices with eigenvalues $i$ and $-i$. If this matrix has rows $(a, b)$ and $(c, d)$ respectively then we have $a + d = 0$ and $ad - bc = 1$ thus $d = -a$ and $a^2 +bc = -1$ or $bc = -(1+a^2)$. Thus you can choose any $a$ and a nonzero $b$, take $d = -a$ and $ c = \dfrac{-(1+a^2)}{b}$. That will give you all required real matrices.

Tony Pizza
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