A zero determinant

Let $A \in \mathcal{M}_n \left( \mathbb{C} \right)$ with $n \geq 2$ such that

$$\det \left ( A+X \right )=\det A + \det X$$

for every matrix $X \in \mathcal{M}_n \left( \mathbb{C} \right)$. Prove that $A=\mathbb{O}$.

Solution

For $X=A$ we will have $2^n\det A = \det(2A) = \det A+ \det A = 2\det A$, forcing $\det A = 0$. Thus the linear application $\mathcal{T}_A$ associated to $A$ has rank $r$ less than $n$. Assume $A\neq \mathbb{O}$, thus $r>0$. It is clear we can build a linear application $\mathcal{T}$ of rank $n-r$ such that $\mathcal{T}_A + \mathcal{T}$ has rank $n$; for that, let $\{b_1,b_2\ldots,b_{n-r}\}$ be a basis of $\ker T_A$, prolong it with $\{f_1,f_2,\ldots,f_r\}$ to a basis of $\mathbb{C}^n$ and define $\mathcal{T}(b_i) = b_i$ and $\mathcal{T}(f_j) = 0$.

Now consider the matrix $X = M(\mathcal{T})$ associated to $\mathcal{T}$; we will have

$$0\neq \det(A+X) = \det(A)+\det(X) = 0+0=0$$

which is clearly a  contradiction