Describe the set of all possible real projector matrices in $\mathbb{R}^{2 \times 2}$

Question

So I started answering my own question as I went but there was a lot of tedious work (for me), and I couldn't find a list of projector matrices, so I figured I would leave this up for posterity. But please let me know if I missed any matrices or if there's an easier way of building the set (and of course if there's any errors). I posted my results as an answer below. Thank you.

Answer 1

Let's look for all rank $1$ projectors, because rank $0$ and rank $2$ are obvious.

If $P$ is such a projector, then its characteristic polynomial is $X^2-X$. In particular, $\det P=0$ and $\operatorname{trace}P=1$. Thus the matrix must have the form $$ P=\begin{bmatrix} a & b \\ c & 1-a \end{bmatrix} $$ If we square it, we get $$ P^2=\begin{bmatrix} a^2+bc & b \\ c & bc+(1-a)^2 \end{bmatrix} $$ so the relations become $$ \begin{cases} a(1-a)-bc = 0 \\[4px] a^2+bc=a \\[4px] bc+(1-a)^2=1-a \end{cases} $$ which however are equivalent to one another. Thus the condition is just $$ bc=a-a^2 $$

Actually, the check about $P$ being idempotent is not necessary. The condition that the trace is $1$ is satisfied and $\det P=0$ implies $bc=a-a^2$. Now the characteristic polynomial is $X^2-X$ and Hamilton-Cayley provides $P^2=P$.

Answer 2

By definition, if $P \in \mathbb{R}^{2 \times 2}$ is a projector matrix then

$$P^2 = P$$

and writing $P$ as

$$P = \left(\begin{array}{cc}a & b\\c & d\end{array}\right)$$

we could solve the system from $P(P - I) = 0$

$$\left(\begin{array}{cc}a & b\\c & d\end{array}\right) \left(\begin{array}{cc}a-1 & b\\c & d-1\end{array}\right) = 0$$

In which case we get the system of equations

$$ \begin{array} aa(a-1) + bc = 0 \\ ab + b(d-1) = 0 \\ c(a-1) + dc = 0 \\ bc + d(d-1) = 0 \end{array} \quad \to \quad \begin{array} aa^2- a + bc = 0 \\ ab + bd - b = 0 \\ ac - c + dc = 0 \\ bc + d^2 - d = 0 \end{array} $$

Subtracting the last equation from the first we get

$$a^2 - d^2 = a - d \implies (a-d)(a+d) = a-d \implies a =d \quad \text{or} \quad a = 1-d$$

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Case 1: $a = 1-d$

Plugging into equation 2 we get

$$ (1-d)b + b(d-1) = 0 $$

Which implies equation 2 is satisfied for all $b$. Plugging into equation 3 we get a similar result for $c$. But multiplying equations 2 and 3 we get a common factor of $bc$, yet the product $eqn. 1 \times eqn. 2$ needs to equal zero. This implies $bc = 0$ (note the remaining factor has no solution which means $bc = 0$ is the only way to satisfy the system). So now, if $b$ and $c$ are arbitrary subject to their product equaling zero we get (eqn 1)

$$ a(a-1) = 0 \implies \boxed{a = 0,1} \implies \boxed{d = 1,0} \quad \text{respectively} $$

our solution set in this case is

$$ P_1 = \left(\begin{array}{cc}1 & b\\c & 0\end{array}\right) \quad \text{and} \quad P_2 = \left(\begin{array}{cc}0 & b\\c & 1\end{array}\right)\,, \quad \text{for all} \, \, b\cdot c = 0 $$

that is, either $b$ or $c$ has to be zero.

For example, $P_2$ with $b = 0$, multiplied by a vector in $\mathbb{R}^2$ scales the $x$ component by factor $c$, adds the result $xc$ to the $y$ component of the vector, and maps the result to the $y$-axis. Here's what it looks like in desmos. Note that $P_2$ becomes an orthogonal projector when $c = 0$.

Here's what the $P_1$ set looks like in desmos (note that the projector is orthogonal when $c = 0$).

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Case 2: $ a = d$

Applied to equation 3

$$c(d-1) + dc = 0 \implies c(2d-1) = 0 \implies c = 0 \quad \text{or} \quad d = \frac{1}{2}$$

Inspecting equation 1 we see that for $c = 0$ we have

$$a(a-1) = 0 \tag{$\star$}$$

A quick analysis of $(\star)$: $a = d = c = 0 \implies b = 0$ which is the trivial case of the zero matrix. If $a = d = 1$ then equation 2 implies $b = 0 = c$ which is the other trivial case of the identity matrix. Done with $(\star)$.

But nontrivially, for $d = 1/2$ we find via elementary algebra that our only restriction is that $bc = 1/4$. That is

$$ P_3 = \left(\begin{array}{cc}1/2 & b\\c & 1/2\end{array}\right) \quad bc = 1/4 $$

It projects onto hyperbolas it looks like, but I'm not sure how to make it work.

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Summary: The only 2D real projector matrices are the identity matrix, the zero matrix, and the sets

$$ P_1 = \left(\begin{array}{cc}1 & b\\c & 0\end{array}\right)\, , \quad P_2 = \left(\begin{array}{cc}0 & b\\c & 1\end{array}\right)\,, \quad \text{for all} \, \, bc = 0\, , \\ \text{and} \quad P_3 = \left(\begin{array}{cc}1/2 & b\\c & 1/2\end{array}\right) \quad bc = 1/4 $$

Answer 3

Hint:

consider that a projector has the eigenvalues $1$ and $0$ ( you can see: Eigenvalues of projection matrix proof). So any projector is represented by a matrix similar to a diagonal matrix with diagonal elements $1$ and $0$.