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 $A^2 = \bf{0}$?


If $A^2=0$, then $A$ is singular. Hence $Au=0$ for some nonzero vector $u$. Pick any vector $v$ that is linearly independent of $u$. Then $Av=au+bv$ for some scalars $a$ and $b$. It follows from $A^2=0$ that
$$
0=A^2v=A(Av)=A(au+bv)=aAu+bAv=bAv=b(au+bv)=abu+b^2v.
$$
Thus $b$ must be zero. Hence $Av=au$ and $Au=0$. Conversely, if $Av=au$ and $Au=0$, it is straightforward to verify that $A^2$ maps both $u$ and $v$ to zero. Thus $A^2=0$.
In short, when $A$ is $2\times2$, $A^2=0$ iff there is some basis $\{u,v\}$ such that $Au=0$ and $Av=au$ for some scalar $a$. In terms of matrices, if $P$ denotes the augmented matrix $[u,v]$, we see that $A^2=0$ iff
$$
A=P\pmatrix{0&a\\ 0&0}P^{-1}
$$
for some nonsingular matrix $P$.
