First I tried to find $A^2$ with
$$ A=\begin{bmatrix} \alpha & \beta\\ \delta & \gamma\\ \end{bmatrix} $$

I multiplied this by itself and got:

$$ \begin{bmatrix} \alpha^2+\beta\delta& \beta(\alpha + \gamma)\\ \delta (\alpha + \gamma) & \delta\beta+\gamma^2\\ \end{bmatrix} $$

I put this in a system:

$$ \left\{ \begin{array}{c} \alpha^2+\beta\delta = 1 \\ \beta(\alpha + \gamma) = 0 \\ \delta (\alpha + \gamma) = 0 \\ \delta\beta+\gamma^2 = 1 \\ \end{array} \right. $$

I tried to solve for $\beta$ first and right away got an issue:

$$\beta = \frac{1-\alpha^2}{\delta}$$

One solution given by my book is: $$ \begin{bmatrix} 1& 0\\ 0 & -1\\ \end{bmatrix} $$

So $\delta$ can be zero but according to my system it can't. How is this possible?

  • 2
    $\begingroup$ When you devided by beta, you assumed it's not 0, since you devided by it. But when it's 0 it's a special case and you need to check what happens there. As you can see there is no contrudiction in beta=0 in the original claims, hence it can be 0. The only thing that yields is that you have two 0s. $\endgroup$
    – Vitali Pom
    Jan 11, 2020 at 15:19
  • 1
    $\begingroup$ You can solve for $\beta$ if $\delta\neq0$, but you have to discuss the system also in the case $\delta=0$. $\endgroup$ Jan 11, 2020 at 15:19
  • 1
    $\begingroup$ According to your system, it can. Your system is not equivalent to the system $$\begin{cases}\beta=\frac{1-\alpha^2}\delta \\ \beta(\alpha + \gamma) = 0\\ \delta (\alpha + \gamma) = 0 \\ \delta\beta+\gamma^2 = 1\end{cases}$$ but rather to the disjunction of systems $$\begin{cases}\delta=0\\ \alpha^2+\beta\delta = 1 \\ \beta(\alpha + \gamma) = 0 \\ \delta (\alpha + \gamma) = 0 \\ \delta\beta+\gamma^2 = 1\end{cases} \lor\begin{cases}\delta\ne0\\ \beta= \frac{1-\alpha^2}\delta \\ \beta(\alpha + \gamma) = 0 \\ \delta (\alpha + \gamma) = 0 \\ \delta\beta+\gamma^2 = 1\end{cases}$$ $\endgroup$
    – user239203
    Jan 11, 2020 at 15:20
  • $\begingroup$ Almost the same question asked 10 hours ago: Curious Case of Idempotent Matrices - Seeking a Generalisation. Is this homework in a course or something? $\endgroup$
    – user856
    Jan 11, 2020 at 15:58
  • $\begingroup$ @Rahul No, just a book I am reading. $\endgroup$ Jan 11, 2020 at 16:00

6 Answers 6


When you solved for $\beta$, you assumed that you could divide through by $\delta$, i.e., you assumed that $\delta \ne 0$. What you needed to do was this:

case 1: $\delta \ne 0$: Then $\beta = \frac{1 - \alpha^2}{\delta}$ ...

Case 2: $\delta = 0$: In this case, we have $\alpha = \pm 1,$ ... (and you fill in the remaining details)

  • $\begingroup$ So I need two systems? $\endgroup$ Jan 11, 2020 at 15:20
  • $\begingroup$ Well...you need two cases for solving your one system; often we have special cases where something might be zero, because dividing by it may or may not be possible. When the thing is zero, often the rest of the system gets much simpler because lots of terms disappear (as in this case), and you can use different methods for solving that simpler system. $\endgroup$ Jan 11, 2020 at 17:08

You made a mistake:

It’s like solving $x=y^2$ and then changing it to $1=\frac{y^2}{x}$ and the saying that $(0,0)$ is not a solution since $x$ cannot be zero.

You assumed $x$ is not zero, the question did not assume anything. You have to solve for the case when $x=0$, same goes for your system.

  • $\begingroup$ This basically gives me a ton of systems doesn't it? $\endgroup$ Jan 11, 2020 at 15:30
  • $\begingroup$ No, you just plug in $\delta =0$ and solve for the remaining values. Because you have considered all the cases from delta is not zero. $\endgroup$ Jan 11, 2020 at 15:36
  • $\begingroup$ I'm trying to solve the second system where $\delta \neq 0$ and got $\begin{cases} \delta \neq 0 \\ \beta=\frac{1-\alpha^2}\delta \\ \alpha + \gamma = 0\\ 1-\alpha^2+\gamma^2 = 1\end{cases}$ = $\begin{cases} \delta \neq 0 \\ \beta=\frac{1-\alpha^2}\delta \\ \alpha = -\gamma\\ \gamma^2 = \alpha^2\end{cases}$ $\\\\$ So $\gamma = \pm \alpha$ and $\alpha = -\gamma$ ? This doesn't make sense $\endgroup$ Jan 11, 2020 at 15:40
  • $\begingroup$ Finished edits. $\endgroup$ Jan 11, 2020 at 15:45
  • $\begingroup$ Your second system does not contain $\delta$, so don't worry about it. Your second system is solved if $\gamma=-\alpha$ or if $\beta=0$. So you have to cases to try for your other three systems. First set $\beta=0$ and see what possibilities are possible for the first equation, then carry those possibilities to your third and fourth systems. Then do it once again with $\gamma=-\alpha$ and $\beta$ could be anything. $\endgroup$ Jan 11, 2020 at 15:52

You assumed $\delta \neq 0$ when you divided by it (when solving for $\beta$). The first system stills valid whether $\delta = 0$ or not.

In fact, when $\delta = 0, |\alpha| = |\gamma| = 1$. If $\gamma = -\alpha, \beta$ can be anything.


Here is one way to find all such matrices $A$ without directly solving for the entries of $A$. Let $\Bbb{K}$ be the base field.

First, suppose that $\Bbb{K}$ is of characteristic unequal to $2$. For a $2\times 2$ matrix $A\neq \pm I$ to satisfy $A^2=I$, we have $(A-I)(A+I)=0$ but $A-I$ and $A+I$ are both non-zero. That is, the dimensions of $\ker(A-I)$ and $\ker(A+I)$ must be both $1$. Therefore, $\ker(A-I)$ and $\ker(A+I)$ is spanned by two non-zero vectors $u$ and $v$. Therefore, if $M$ denotes the matrix $$M=\begin{bmatrix}\vert &\vert\\ u &v\\\vert&\vert\end{bmatrix}.$$ Then, $A=MJM^{-1}$, where $$J=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$ If $\Bbb K=\Bbb Q$, then we can take $u=\begin{bmatrix}\frac{2r}{1+r^2}\\\frac{1-r^2}{1+r^2}\end{bmatrix}$ and $v=\begin{bmatrix}\frac{2s}{1+s^2}\\\frac{1-s^2}{1+s^2}\end{bmatrix}$ with $r,s\in\left(-1,1\right]\cap\Bbb Q$ and $r\neq s$, so that $$A=\frac{1}{(r-s)(1+rs)}\begin{bmatrix}(r+s)(1-rs)&-4rs\\(1-r^2)(1-s^2)&-(r+s)(1-rs)\end{bmatrix}.$$ If $\Bbb K=\Bbb R$, then we can take $u=\begin{bmatrix}\cos x\\\sin x\end{bmatrix}$ and $v=\begin{bmatrix}\cos y\\\sin y\end{bmatrix}$ with $x,y\in\left[0,\pi\right)$ and $x\neq y$, so that $$A=\frac{1}{\sin(x-y)}\begin{bmatrix}-\sin(x+y)&2\cos x\cos y\\-2\sin x\sin y&\sin(x+y)\end{bmatrix}.$$ If $\Bbb K=\Bbb C$, then we can take $u=\begin{bmatrix}e^{i\lambda}\cos x \\\sin x\end{bmatrix}$ and $v=\begin{bmatrix}e^{i\mu}\cos y \\\sin y\end{bmatrix}$ with $x,y\in\left[0,\pi\right)$ and $\lambda,\mu\in[0,2\pi)$ such that

  • if $x=0$ or $x=\pi/2$, then $\lambda=0$;
  • if $y=0$ or $y=\pi/2$, then $\mu=0$;
  • either $x\neq y$, or $x=y$ and $\lambda \neq \mu$.

In this case, $$A=\frac{1}{\cos\frac{\lambda-\mu}{2}\sin(x-y)-i\sin\frac{\lambda-\mu}{2}\sin(x+y)}\begin{bmatrix}-\cos\frac{\lambda-\mu}{2}\sin(x+y)+i\sin\frac{\lambda-\mu}{2}\sin(x-y)&2e^{i\left(\frac{\lambda+\mu}{2}\right)}\cos x\cos y \\2e^{-i\left(\frac{\lambda+\mu}{2}\right)}\sin x\sin y &\cos\frac{\lambda-\mu}{2}\sin(x+y)-i\sin\frac{\lambda-\mu}{2}\sin(x-y)\end{bmatrix}$$

Note that the set of such matrices $A$ is in a $1$-to-$1$ correspondence with the right-coset space of $\operatorname{GL}_2(\Bbb{K})$ modulo the subgroup of diagonal matrices (isomorphic to $\Bbb{K}^\times \times \Bbb{K}^\times$). Particularly, if $\mathbb{K}$ is a finite field with $q$ elements, then there are exactly $$\frac{(q^2-1)(q^2-q)}{(q-1)(q-1)}=q(q+1)=q^2+q$$ such matrices $A$. (For instance, when $q=3$, there are $12$ possible choices of $A$: $\pm\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, $\pm\begin{bmatrix}0&1\\1&0\end{bmatrix}$, $\pm\begin{bmatrix}1&1\\0&-1\end{bmatrix}$, $\pm\begin{bmatrix}1&-1\\0&-1\end{bmatrix}$, $\pm\begin{bmatrix}1&0\\1&-1\end{bmatrix}$, $\pm\begin{bmatrix}1&0\\-1&-1\end{bmatrix}$.)

In characteristic $2$, we note that $A^2-I=(A-I)^2$. Since $A\neq I$, the Jordan canonical form of $A$ is $$J=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$$ We start with an arbitrary ordered basis $(u,v)$ of $\Bbb{K}^2$, and then declare that $(A-I)v=u$. That is, with $$M=\begin{bmatrix}\vert &\vert\\ u &v\\\vert&\vert\end{bmatrix},$$ we have $A=MJM^{-1}$.

Note that the set of such matrices $A$ is in a $1$-to-$1$ correspondence with the right-coset space of $\operatorname{GL}_2(\Bbb{K})$ modulo the subgroup of upper-diagonal matrices with identical diagonal entries (isomorphic to $\Bbb{K}^\times \times\Bbb K$). Particularly, if $\mathbb{K}$ is a finite field with $q$ elements, then there are exactly $$\frac{(q^2-1)(q^2-q)}{(q-1)q}=(q+1)(q-1)=q^2-1$$ such matrices $A$. (For instance, when $q=2$, there are only three possibilities for $A$: $\begin{bmatrix}0&1\\1&0\end{bmatrix}$, $\begin{bmatrix}1&1\\0&1\end{bmatrix}$, and $\begin{bmatrix}1&0\\1&1\end{bmatrix}$.)

In general, all possible $A$ (regardless of the characteristic of $\Bbb K$) are given by the two parametrizations below.

  • $A=\begin{bmatrix}\alpha&\beta\\\frac{1-\alpha^2}{\beta}&-\alpha \end{bmatrix}$ with $\alpha\in \Bbb{K}$ and $\beta\in \Bbb K\setminus\{0\}$.
  • $A=\begin{bmatrix}\alpha&0\\\delta&-\alpha \end{bmatrix}$ with $\alpha=\pm 1$ and $\delta\in\Bbb K$. (If $\Bbb K$ has characteristic $2$, then $\alpha=1$ and $\delta$ must be taken from $\Bbb K\setminus\{0\}$.)
  • $\begingroup$ I have no idea what most of that means but thanks $\endgroup$ Jan 11, 2020 at 15:45
  • $\begingroup$ This is... unnecessary. $\endgroup$ Jan 11, 2020 at 16:53

I know, because $\alpha ^2+\beta\delta =1$, so you think $\beta=\frac{1-\alpha ^2}{\delta}$

But for $\alpha^2+\beta\delta =1$, you can only know $\beta\delta=1-\alpha ^2$.

when $\delta =0$, now $0=1-\alpha ^2$;

when $\delta\neq 0$, now $\beta=\frac{\beta\delta}{\delta}=\frac{1-\alpha ^2}{\delta}$


I found a much simpler way to solve this (probably the one intended by the authors).

Since $$AA = I_2 \Leftrightarrow A = I_2.A^{-1} \Leftrightarrow A = A^{-1}$$

So, $$A=\begin{bmatrix} a & b\\ c & d\\ \end{bmatrix} = A^{-1} = \det(A)^{-1}*\begin{bmatrix} d & -b\\ -c & a\\ \end{bmatrix} $$

$\det(A)^{-1} = \frac{1}{ad-bc} = \delta$

This means that $A=A^{-1}$ equals $$\begin{bmatrix} d\delta & -b\delta\\ -c\delta & a\delta\\ \end{bmatrix}$$

Putting this all in a system: $$ \left\{ \begin{array}{c} a = d\delta \\ b = -b\delta \\ c = -c\delta \\ d = a\delta \\ \end{array} \right. $$

Now there's 2 possibilities: $b=c=0$ or $b,c \neq 0$

Starting with $b=c=0$:

$$ \left\{ \begin{array}{c} b = c = 0 \\ a = d\delta \\ d = d\delta^2 \Leftrightarrow \pm 1 = \delta \\ \end{array} \right. $$

Now for the values of $\delta$ we get:

$\frac{1}{ad} = 1 \lor \frac{1}{ad} = -1 \Leftrightarrow a = 1/d \lor a = -1/d$

$$ \left\{ \begin{array}{c} b = c = 0 \\ a = 1/d \\ d \in \mathbb{R}\\ \end{array} \right. $$ $$ \left\{ \begin{array}{c} b = c = 0 \\ a = -1/d \\ d \in \mathbb{R}\\ \end{array} \right. $$

Now for the case of $b,c \neq 0$ we have:

$$\left\{ \begin{array}{c} a = -d \\ \delta=-1 \\ \end{array} \right.$$

Solving for the value of d in this case:

$$\delta = -1 \Leftrightarrow \frac{1}{ad-bc} \Leftrightarrow d^2+bc = 1 \Leftrightarrow \\ \pm d = \sqrt{1-bc}$$

Finally the system becomes:

$$\left\{ \begin{array}{c} a = -d \\ d = \mp \sqrt{1-bc} \\ b,c \in \mathbb{R} \setminus \{0\} \end{array} \right.$$

I think that covers them all... how'd I do?


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