Determine all $2\times2$ matrices $A$ such that $A^2=0$. This is from Lang's introduction to Linear Algebra page no 61.

Determine all $2\times 2$ matrices $A$ such that $A^2 = 0$.

Let $A=\begin{pmatrix}
   a & b \\
   c & d
   \end{pmatrix}$
$A^2=\begin{pmatrix}
   a^2+bc & ab+bd \\
   ac+cd & d^2+cb
   \end{pmatrix}$
Equating all the four terms of the matrix to zero and solving,
$a^2+bc=0$ and $d^2+bc=0$ gives $a=\pm d$
Solving $ab+bd=0$ and $ac+cd=0$ gives $a=-d$ OR $b=c$, which gives $a=d=0$
My question: I know that if $A$ is an $n\times n$ matrix with all the diagonal elements and all the elements below it equal to zero then $A^n=0$, so in this question
 \begin{equation}
   \begin{pmatrix}
   0 & x \\
   0 & 0
   \end{pmatrix}
   \end{equation} qualifies as $A$, but it does not agree with $b=c$ as in above solution, so where did I go wrong?
And if I go with $b=c$, then $A^2$ does not equal $0$.
 A: Well you have your 4 equations:
$$\cases{a^2+bc=0 \\ ab+bd=0 \\ ac+cd=0 \\ d^2+bc=0}$$
And you know that if $A^2=0$ then $det(A^2)=det(A)^2=0$ and thus $det(A)=0$. So you have a fifth equation:
$$ad-bc=0 \Rightarrow ad=bc$$
Now replace this in your system and factorize it:
$$\cases{a^2+ad=a(a+d)=0 \\ ab+bd=b(a+d)=0 \\ ac+cd=c(a+d)=0 \\ d^2+ad=d(a+d)=0 \\ ad=bc}$$


*

*If $a+d\neq0$ then $a=b=c=d=0$

*If $a+d=0$ then all these equations are equivalent and you're left with two equations (you already know you'll probably have 2 degrees of liberty):


$$\cases{a+d=0 \\ ad=bc}\Rightarrow\cases{d=-a \\ -a^2=bc}$$


*

*If $b=0$ then $a=d=0$ and you're just left with $c\in\Bbb{R}$

*If $b\neq0$ then $\cases{d=-a \\ c=-\frac{a^2}{b}}$


Now this means that you have two possible matrices s.t. $A^2=0$:
$A=\left(\begin{array}{cc} a && b \\ -\frac{a^2}{b} && -a \end{array}\right),(a,b)\in\Bbb{R}\times\Bbb{R}^*$ or $A=\left(\begin{array}{cc} 0 && 0 \\ c && 0 \end{array}\right),c\in\Bbb{R}$
You also have their transposes $\left((A^T)^2=A^TA^T=(AA)^T=(A^2)^T=0^T=0\right)$
A: When you take the difference of 2 equations, it does not mean that solutions to the new equation will satisfy both of the old equations.
For example, if $a=d=0$, then this satisfies your conclusion of $a= \pm d$, but doesn't always satisfy your initial conditions of $a^2+bc = 0, d^2 + bc = 0$ (in particular if $b=c\neq0$).
You can read Proof that 0=1 to see another example of how this method of solving equations can go wrong.
