A question about the determinant of traceless matrices I have a problem in proving the following lemma. I appreciate any help. Thanks in advance.

Lemma. Let matrix $A \in \Bbb C^{n \times n}$ be traceless. If $$\det(A+C)=0$$ for any traceless matrix $C$ with $\det(C)=0$, then $A=0$.

 A: Suppose that $A\ne  0$. We can assume that the size $n>1$.  We can also assume that $A$ is in the Jordan normal form.
If at least one diagonal entry is not $0$, $n\ge 3$, we can add to $A$ a traceless diagonal matrix with det $0$ to make  matrix still upper triangular and with all diagonal entries nonzero, a contradiction. If $n=2$, then the matrix is $\begin{pmatrix} a & 0\\ 0 & -a\end{pmatrix}$, $a\ne 0$,  since $\operatorname{tr}(A)=0$ and $A$ is in Jordan form, a contradiction because
$\det(A)=-a^2\ne 0$.
So all diagonal entries are zero. If $A$ has only one Jordan block, we can add matrix unit $E_{n,1}$ (the matrix with $(n,1)$ entry $1$ and zeroes everywhere else) and get a contradiction. If the number of blocks is more than $1$ (the number of blocks cannot be $0$ or $n$ since $A$ is not $0$), we can add a few matrix units to make the number of blocks $1$ and then add $E_{n,1}$ to make the determinant $\ne 0$. The total matrix added will be traceless and with det $0$.
A: The following proof works over any field. The condition that $\operatorname{tr}(A)=0$ is not needed.
We may assume that $A$ is in rational canonical form, i.e. $A=\lambda I_m\oplus M_1\oplus\cdots\oplus M_k$ where each $M_i$ is a companion matrix of the form $\pmatrix{0&a\\ I&v}$ and of size at least $2\times2$. There are five possibilities:

*

*$A=M_1\oplus\cdots\oplus M_k$. For each $i$, choose a square matrix $C_i$ of the same size as $M_i$ such that its only possibly nonzero entry is the element at the top right corner and $M_i+C_i$ is nonsingular. Then $C=C_1\oplus\cdots\oplus C_k$ is traceless and singular but $A+C$ is nonsingular. Hence we arrive at a contradiction and this case is impossible.

*$A=\lambda I_m\oplus M_1\oplus\cdots\oplus M_k$ where $\lambda\ne0$. Contradiction arises if we pick $C=0_{m\times m}\oplus C_1\oplus\cdots\oplus C_k$.

*$A=0_{m\times m}\oplus M_1\oplus\cdots\oplus M_k$ for some $m>1$. Contradiction arises if we pick $C=P\oplus C_1\oplus\cdots\oplus C_k$, where $P$ is any $m\times m$ permutation matrix with a zero diagonal.

*$A=0_{1\times 1}\oplus M_1\oplus\cdots\oplus M_k$. Suppose $M_1$ is $r\times r$. Then contradiction arises if we pick $C=(E_{21}+E_{1,r+1})\oplus C_2\cdots\oplus C_k$, where $E_{ij}$ denotes the $(r+1)\times(r+1)$ matrix whose only nonzero entry is a $1$ at the $(i,j)$-th position. Note that $(0_{1\times1}\oplus M_1)+(E_{21}+E_{1,r+1})$ is nonsingular because it is in the form of
$$
\pmatrix{0&0&1\\ 1&0&a\\ 0&I_{r-1}&v}.
$$

*$A=\lambda I$. By taking $C=0$, we see that $A$ is singular. Hence $\lambda$ must be zero and $A=0$.

