If $A$ is a rank $1$ matrix, then $A^2= \operatorname{Tr}A \cdot A$ I am posting this question because I want to know if my proof is correct (I know that the result holds for $\mathbb{K}=\mathbb{C}$ and I see no reason why it wouldn't work for an arbitrary field, but I just want to be sure).
Claim : Let $A \in \mathcal{M}_n(\mathbb{K})$ ($\mathbb{K}$ is a field, $n\in \mathbb{N}, n\ge 2$) such that $\operatorname{rank}A=1$. Then we have that $A^2=\operatorname{Tr}A\cdot A$.
Proof : Since $\operatorname{rank}A=1$, $A's$ lines are proportional i.e.
$A= \begin{pmatrix}
b_1c_1 & b_1c_2 &...& b_1c_n\\
b_2c_1 & b_2c_2 &...& b_2c_n\\
... & ... & ...& ...\\
b_nc_1 & b_nc_2 &...& b_nc_n\\
\end{pmatrix}=\begin{pmatrix}
b_1 & 0 &...& 0\\
b_2 & 0 &...& 0\\
... & ... & ...& ...\\
b_n & 0 &...& 0\\
\end{pmatrix}\cdot \begin{pmatrix}
c_1 & c_2 &...& c_n\\
0 & 0 &...& 0\\
... & ... & ...& ...\\
0 & 0 &...& 0\\
\end{pmatrix}.$
Let $B:=\begin{pmatrix}
b_1 & 0 &...& 0\\
b_2 & 0 &...& 0\\
... & ... & ...& ...\\
b_n & 0 &...& 0\\
\end{pmatrix}$ and $C:=\begin{pmatrix}
c_1 & c_2 &...& c_n\\
0 & 0 &...& 0\\
... & ... & ...& ...\\
0 & 0 &...& 0\\
\end{pmatrix}$.
We have that $A^2=B(CB)C=\operatorname{Tr}A\cdot BC=\operatorname{Tr}A\cdot A$ and we are done.

It's technically correct, but I think it's clearer to write $A$ as $bc^T$ with $b,c\in\mathbb K^n$. The identity can then be proved easily as
$$
A^2=(bc^T)(bc^T)=b(c^Tb)c^T=(c^Tb)bc^T=(\operatorname{Tr}A)A.
$$
If you decompose $A$ into a product of two square matrices $B$ and $C$, one may not immediately see why $B(CB)C=(\operatorname{Tr}A)BC$.
