<s>[INST] Find all matrices $A\in \mathbb{R}^{2\times2}$ such that $A^2=\bf{0}$ Attempt:
Let's consider $A = 
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$.
$$\begin{align}
A^2 &= 
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\cdot
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \\&=
\begin{bmatrix}
a\cdot a+b\cdot c & a\cdot b + b\cdot d \\
c\cdot a + d\cdot c & c\cdot b + d\cdot d 
\end{bmatrix}\\ &= \bf{0}
\end{align}$$
This gives us the system of equations:
$$\begin{align}
a\cdot a+b\cdot c &= 0 \tag{1}\\
a\cdot b + b\cdot d  = b\cdot(a+d)&= 0 \tag{2}\\
c\cdot a + d\cdot c = c\cdot(a+d)&= 0 \tag{3}\\
c\cdot b + d\cdot d &= 0 \tag{4}
\end{align}$$
Now from equation $(2)$ and $(3)$, we have eight cases: 


*

*$b = 0$

*$c = 0$

*$a+d = 0$
and 5 combinations of (1,2,3) which I won't bother listing.
Case 1 ($b=0$):
$b=0$ implies $a = 0$ in equation $(1)$ and $d = 0$ in equation $(4)$. This means that if $A = \begin{bmatrix}0&0\\c&0\end{bmatrix}$ then $A^2=\bf{0}$.
Case 2 ($c=0$):
From symmetry with $b$, $c=0 \implies A=\begin{bmatrix}0&b\\0&0\end{bmatrix}$.
So, we only consider cases where $b\neq0$ and $c\neq 0 $ which leaves us only with case 3 ($a+d=0$).
Case 3 ($a+d=0$):
In equation (1), $a+d=0 \implies a\cdot d - b\cdot c = 0$. So $A=\begin{bmatrix}a&b\\c&-a\end{bmatrix}$ is not invertible, $A^2 = \bf{0}$.
In summary, if $A$ has one of the following forms:
$$\begin{bmatrix}0&b\\0&0\end{bmatrix},
\begin{bmatrix}0&0\\c&0\end{bmatrix},
\begin{bmatrix}a&b\\c&-a\end{bmatrix} \text{ (and not invertible) }$$
then $A^2=\bf{0}$.

Questions:


*

*Is this a correct proof?

*What is the standard proof?


"Strange question":


*

*How can I know if there was only the 8 cases? As in, how do I know that only these 8 cases are relevant to $A^2 = \bf{0}$?

 [/INST] Your proof is correct, although rather naive---in the sense that you've converted the matrix equation into a system of simultaneous equations and solved it. There's no "standard" proof per se, but there are many more conceptual ways to prove it.
One example is the eigenvalue argument provided by Bungo in the comments. Another way is to note that if $A^2=0$, then the minimal polynomial of $A$ must be $x^2$ (except the trivial case $A=0$), and since the minimal polynomial of a matrix divides its characteristic polynomial, then the characteristic polynomial of $A$ must be $kx^2$ for a constant $k$. It is easy to deduce all the possible forms of $A$ from here on.
The advantage of these more "high-level" arguments is that they generalise easily, to e.g. higher dimensional vector spaces.
</s>