I am taking linear algebra for the first time and am struggling with the concept of idempotent matrices. I know that $A = A^2$ is the concept behind it, but I can't seem to understand HOW one would find the entries, and the explanations given confused me quite a lot. I was hoping someone could give me the gist of this concept and point me in the right direction.
For example, I am faced with the question of "Find all $2\times 2$ matrices such that $A^2=A$"
Currently, I know that with the matrix $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ that:
$a = a^2 + bc\\
b = ab + bd\\
c = ca + cd\\
d = bc + d^2$
but I'm now unsure of how to determine entries off of that.
Best Answer
Consider the system $$a = a^2 + bc \tag{1}$$ $$b = ab + bd \tag{2}$$ $$c = ca + cd \tag{3}$$ $$d = bc + d^2 \tag{4}$$ Then, $(1) - (4)$ produces $$a - d = a^2 - d^2 = (a - d)(a + d),$$ so $a = d$ or $a + d = 1$.
Assume, as one possible case, $a + d = 1$. Note that this implies $(2)$ and $(3)$ are automatically satisfied. As expected, if you substitute $d = 1 - a$ into $(4)$, you just get back $(1)$, so we now have only two equations (in this case): \begin{align*} bc &= a - a^2 \\ d &= 1 - a. \end{align*} We should let $a$ be a free variable; no matter what value $a$ takes, there will always be a solution, from which we can uniquely determine $d$. If $a - a^2 \neq 0$, then we can also let $b$ be a free variable, with the caveat that $b \neq 0$, and $c = \frac{a - a^2}{b}$. This produces a family of solutions: $$\fbox{$\begin{pmatrix} a & b \\ \frac{a - a^2}{b} & 1 - a \end{pmatrix}, \qquad a, b \in \Bbb{R}, \quad b \neq 0$}.$$
If $a - a^2 = 0$, i.e. $a = 0$ or $a = 1$ (and $d = 1$ or $d = 0$ respectively), then we have to be more careful. We would have two cases: $b = 0$ or $c = 0$. This gives us four families of solutions (which, as you should verify, are all solutions: $$\fbox{$\begin{pmatrix} 1 & b \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ c & 0 \end{pmatrix}, \begin{pmatrix} 0 & b \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ c & 1 \end{pmatrix}, \qquad b, c \in \Bbb{R}$}.$$
This exhausts all the cases where $a + d = 1$. Otherwise, $a = d$. Then $(2)$ becomes $$b = 2ab$$ i.e. $b = 0$ or $a = d = \frac{1}{2}$. Since $a = d = \frac{1}{2}$ is covered by the $a + d = 1$ case, we dismiss it. So, we assume $b = 0$. Similar analysis of $(3)$ shows us that $c = 0$. Hence, $(1)$ and $(4)$, along with our assumption that $a = d$, tells us that $a = d = 0$ or $a = d = 1$. That is, we have just two remaining solutions:
$$\fbox{$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$},$$ i.e. the zero and identity matrices.