Find $A \in\mathcal M_2(\mathbb{R})$ such that $A^2 =A$

Question

I have a question: find $A \in\mathcal M_2(\mathbb{R})$ such that $A^2=A$. I came up with these four equations:

$$\begin{cases}a^2+bc=a\\ab+bd=b\\ac+cd=c\\bc+d^2=d\end{cases}$$

Can you guide me how to solve this system of equations?

Answer 1

$A:=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ is such that $A^2=A$. This is the matrix of the linear projection $p:\mathbb R ^2 \to \mathbb R ^2, (x,y)\mapsto (x,0)$. It is the projection on $\text{Im } p=\mathbb R(1;0)$ parallel to $\text{Ker }p=\mathbb R(0,1)$. Generally speaking, a matrix $A$ such that $A^2=A$ is the matrix of a linear projection $p:\mathbb R ^2 \to \mathbb R ^2(p^2=p)$. This is a classic result of linear algebra. I don't think it's helpful to go through the proposed equations.

Answer 2

How to solve your system of equations:

$b(a+d-1)=0$ means $b=0$ or $a+d-1=0.$

$c(a+d-1)=0$ means $c=0$ or $a+d-1=0.$

So $d=1-a$, with $bc=a-a^2$, where $a$ is arbitrary,

or $b=0$ and $c=0,$ so $a^2=a$ and $d^2=d$, so $a,d\in\{0,1\}$.

Answer 3

Following same reasoning in this answer, We need $$ \begin{align} d &= 1-a, \\ ad &= bc \end{align} $$

Choose $a=1, b=1$ and $b\neq 0$, we get $c=0$ and $d=0$.