What is a two parameter family of matrices? Let $A\in \mathbb{C}^{m\times n}$ and $B\in\mathbb{C}^{n\times m}$ exist such that $$ABA=A \text{ and } BAB=B.$$ I am to show that all $B$ that satisfy the above form a two-parameter family of matrices, given that
$$A=\begin{bmatrix}0 & 1\\ 0 & 0 \end{bmatrix}.$$ My first question is what is a two parameter family of matrices? I found that if 
$$B=\begin{bmatrix}a & b\\ c & d \end{bmatrix},$$
then $ad = b$ and $c=1$ (from the definition of $B$ above). How can I get this into a form that the question is asking?
My guess (this could be wrong if my definition of two-parameter family is not right) is that given $a,d\in\mathbb{C}$, I can form $B$ by
$$\begin{bmatrix}0 & 0\\ 1 & 0 \end{bmatrix} + a\begin{bmatrix}1 & 0\\ 0 & 0 \end{bmatrix} + d\begin{bmatrix}0 & 0\\ 0 & 1 \end{bmatrix} + ad\begin{bmatrix}0 & 1\\ 0 & 0 \end{bmatrix}.$$
That is, I created a sort of mapping such that given two parameters, I can form any $B$. Is this correct?
Thanks
 A: We are given $A$ and have to solve the system
$$ABA=A\quad\wedge\quad BAB=B\tag{1}$$
for $B$. You have done the calculations (which I have not checked), and found out that $(1)$ is equivalent with
$$ad=b\quad\wedge\quad c=1\ .\tag{2}$$
This means that we can choose $a$ and $d$ arbitrarily, then $b$ and $c$ are determined by $(2)$. The solution set of $(1)$ therefore is given by
$$\left\{\left[\matrix{a&ad \cr1&d\cr}\right]\biggm|\> a, d\in{\mathbb C}\right\}\ .$$
Of course you can present this set officially as a family $\bigl(B_{(a,d)}\bigr)_{(a,d)\in{\mathbb C}^2}\>$, whereby
$$B_{(a,d)}:=\left[\matrix{a&ad \cr1&d\cr}\right]\ .$$
A: When $A\in M_{m,n}$ is given, there always exists some $B\in M_{m,n}$ satisfying $ABA=A,BAB=B$. When $m=n$ and $A$ is generic (that implies that $A$ is invertible), $B=A^{-1}$ is unique; however, in particular cases, like that of the OP, $B$ depends on some parameters. 
When $m\not= n$ and $A$ is generic, there is always an infinity of solutions $B$. For example, if $m=5,n=3$, then $B$ depends on $6$ parameters. 
Moore, Penrose had the great idea to add the following two conditions to get the uniqueness of $B$:
$AB$ and $BA$ are hermitian (or symmetric over $\mathbb{R}$); cf.
https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse
