Given a traceless matrix $C\in M\_n(\mathbb{F})$, i.e., tr$(C)=0$, what is the relationship between tr$|\mathbb{I}+C|$ and tr$|C|$? The two matrices are of dimension $n$.



Matrix inequality between a traceless matrix and identity
I will assume that $C$ is hermitian, as you say in the comments. I will also assume $n\ge2$.


Summary
-------


There are three cases, depending on the value of $\mathrm{tr}|C|$:


1. For $\mathrm{tr}|C|\le2(n-2)$, we have
$$
0\le \mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ .
$$
2. For $2(n-1)\ge\mathrm{tr}|C|>2(n-2)$, there are two possibilities: 
$$
4-n+\mathrm{tr}|C|\le\mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|
$$
or 
$$\mathrm{tr}|\mathbb I+C|=n\ .$$
3. For $\mathrm{tr}|C|>2(n-1)$, we have
$$
4-n+\mathrm{tr}|C|\le\mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ .
$$




---


Derivation
----------


We can decompose $C$ into eigenspaces with positive eigenvalues, eigenvalues in $[-1;0]$, and those below $-1$, respectively, 
$$
C=P-R-M\ ,
$$
where $P\ge0$ (positive semi-definite), $0\le R\le\mathbb{I}$, and $M\ge\mathbb{I}$, respectively, and $P$, $R$, and $M$ have orthogonal supports. Then,
$$
0=\mathrm{tr}\,C=\mathrm{tr}\,P-\mathrm{tr}\,R-\mathrm{tr}\,M\ ,
$$
and hence $\gamma:=\mathrm{tr}\,P=\mathrm{tr}\,M+\mathrm{tr}\,R$. Let us further denote $s:=\mathrm{tr}\,R$, and then $0\le\mathrm{tr}\,M=\gamma-s$, ie., there is a constraint $\gamma\ge s$.


What are the constraints on $\gamma$ and $s$? Generally, $\gamma\ge0$ and $s\in[0;n]$. Note, however, that for $s> n-1$, $R$ must have full rank and thus $\gamma=0$. This is in contradiction to $\gamma\ge s$, ruling out $s>n-1$. Further, if $n-1\ge s> n-2$, then $\gamma\ge s>0$, and $P-R$ have rank $n$, i.e., $M=0$, and thus $\gamma=s$. It is easy to see that there are no further constraints by choosing $P$ and $M$ rank one if $s\le n-2$.


As we will now see, $\gamma$ and $s$ fully characterize both $\mathrm{tr}\,|C|$ and $\mathrm{tr}\,|\mathbb I+C|$.


First, $|C|=P+R+M$, and thus 
$$
\mathrm{tr}|C|=\gamma+s+(\gamma-s)=2\gamma\ .
$$


On the other hand, $|\mathbb I+C|=(\mathbb I+P) + (\mathbb I-R)+(M-\mathbb I)$, and thus 
\begin{align}
\mathrm{tr}|\mathbb I+C| &= (n+\gamma) +(n-s)+((\gamma-s)-n) \\
&= n + 2\gamma-2s\\
&=(n-2s)+\mathrm{tr}|C|\ .
\end{align}


We now have to see what the possible choices for $s$ are, given $\mathrm{tr}|C|=2\gamma$. We must always have $0\le2s\le \mathrm{tr}|C|$. 


1. For $\mathrm{tr}|C|\le2(n-2)$, any $0\le 2s\le\mathrm{tr}|C|$ is admissible, giving the general constraint
$$
0\le \mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ .
$$
2. For $2(n-1)\ge\mathrm{tr}|C|>2(n-2)$, we have two possibilities: $0\le s\le n-2$ or $2s=\mathrm{tr}|C|$. This gives the two possibilities
$$
4-n+\mathrm{tr}|C|\le\mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|
$$
or 
$$\mathrm{tr}|\mathbb I+C|=n\ .$$
3. Finally, for $\mathrm{tr}|C|>2(n-1)$, we have that $0\le s \le n-2$, and thus only have the solution
$$
4-n+\mathrm{tr}|C|\le\mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ .
$$


