# Determine all $2 \times 2$ real matrices $A$ such that $(1) \ \ A^2=I$, $(2) \ \ A^2=0$

Determine all $$2 \times 2$$ real matrices $$A$$ such that $$(1) \ \ A^2=I$$, $$(2) \ \ A^2=0$$

I came across this problem recently where I have to determine all the $$2\times2$$ matrices satisfying the aforementioned criteria. The problem is that the equations which I get after multiplication are not so easy to work with and I'm unable to proceed.

If someone could help me with these equations and solve for the various cases, that'd be really helpful.

Answer to part one: $$A=I$$ $$A=-I$$ $$a_{11}=a_{22}=0 \ , \ a_{12}a_{21}=1$$$$a_{11}=-a_{22}\neq0, \ a_{11}^2+a_{12}a_{21}=1$$

Answer to part two: $$a_{11}=a_{12}=a_{22}=0$$$$a_{11}=a_{21}=a_{22}=0$$$$a_{11}=-a_{22}\neq0, \ a_{11}^2+a_{12}a_{21}=0$$

• How are you getting these equations? I can't see what they have to do with the question. – saulspatz Mar 17 at 13:58
• @saulspatz those are the actual answers given in the textbook. I'm looking for a way to reach them. – s0ulr3aper07 Mar 17 at 14:44

Case $$A^{2} = I$$.

Two possibilities are of course $$A = \pm I$$. If $$A \ne \pm I$$, then the characteristic polynomial of $$A$$ is $$x^{2} - 1$$, which means $$A$$ has trace $$0$$ and determinant $$-1$$. Therefore $$\begin{equation*} \begin{bmatrix} a & b\\ c & -a \end{bmatrix}, \end{equation*}$$ where $$a^{2} + b c = 1$$.

Case $$A^{2} = 0$$.

One possibility is of course $$A = 0$$. If $$A \ne 0$$, then the characteristic polynomial of $$A$$ is $$x^{2}$$, which means $$A$$ has trace and determinant $$0$$. Therefore $$\begin{equation*} \begin{bmatrix} a & b\\ c & -a \end{bmatrix}, \end{equation*}$$ where $$a^{2} + b c = 0$$, and $$a, b, c$$ are not all zero.

In part one, distinguish now

• the case when $$a = \pm 1$$, and then either $$b = 0$$ and $$b$$ is arbitrary, or vice versa;
• the case when $$a \ne \pm 1$$, and then $$b \ne 0$$, say, can be chosen arbitrarily, and $$c = (1 - a^{2})/b \ne 0$$.

In part two, distinguish now

• the case when $$a = 0$$, and then either $$b = 0$$ and $$c$$ is arbitrary, or vice versa;
• the case when $$a \ne 0$$, and then $$b \ne 0$$, say, can be chosen arbitrarily, and $$c = -a^{2}/b\ne 0$$.
• Thanks a lot! But I have not yet learnt about characteristic polynomial, so if it's not too much to ask for, could you please add some information about it and how you've used it in this context? – s0ulr3aper07 Mar 17 at 15:04
• @s0ulr3aper07 have you been taught about eigenvalues? – Andreas Caranti Mar 17 at 16:21
• Yes, I have learnt about eigenvalues and have come across the characteristic polynomial while doing so. The way you've used it to find trace and determinant is new to me. – s0ulr3aper07 Mar 17 at 16:25
• @s0ulr3aper07, ok, so I believe you just need to know a couple of things. The first one is that the characteristic polynomial $A - x I$ of $2 \times 2$ matrix $A$ has the form $x^{2} - T x + D$, where $T$ is the trace of $A$, and $D$ is its determinant. It is just a matter of expanding $\det\left( \begin{bmatrix}a - x&b\\c&d-x\end{bmatrix}\right)$, and then... – Andreas Caranti Mar 17 at 17:01
• The second thing is to note that if $A^{2} = I$, and the eigenvalues of $A$ are both $1$, then $A = I$, while if they are both $-1$, then $A = -I$. Just check that if this is not the case, then $A^{2} \ne I$. – Andreas Caranti Mar 17 at 17:04

$$A = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \Rightarrow A^2 = \begin{bmatrix} a^2+bc & (a+d)b\\ (a+d)c & bc +d^2 \end{bmatrix}$$

In both cases, you have to impose $$(a+d)b = (a+d)c = 0$$.

1. Case $$a+d \neq 0$$: then $$b = c = 0$$ and $$A^2 = \begin{bmatrix} a^2 & 0\\ 0 & d^2 \end{bmatrix}$$ and it's easy to continue.
2. Case $$a+d = 0$$: $$A^2 = \begin{bmatrix} a^2+bc & 0\\ 0 & bc + d^2 \end{bmatrix}$$ and it's easy to continue.