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 $n$.
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1$\begingroup$ What does $|C|$ mean? $\endgroup$– Fedor PetrovCommented Feb 5, 2016 at 7:35
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$\begingroup$ $C$ is a traceless Hermitian matrix, and $|C|$ is simply $\sqrt{C^{\dagger}C}$. Since $C$ is traceless, we know that tr$|C|$ is just the sum of the absolute values of the eigenvalues, which makes tr$|C|>0$. It will only be zero if $C=0$ and it is not. $\endgroup$– JanusCommented Feb 5, 2016 at 7:40
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1$\begingroup$ ... and what is $\Bbb{F}$? $\endgroup$– abxCommented Feb 5, 2016 at 7:56
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3$\begingroup$ At first, 'arbitrary field' and 'either $\mathbb{R}$ or $\mathbb{C}$' are different things. At second, it is a bad idea to denote the set of Hermitian matrices by $M_n(\mathbb{C})$. At third, for Hermitian matrix singular values are just absolute values of eigenvalues, thus we have to say something about $A=\sum |c_i|$ and $B=\sum |1+c_i|$, where $c_i$ are eigenvalues of $C$ and $\sum c_i=0$. It is an excersise to describe the locus of possible points $(A,B)$. $\endgroup$– Fedor PetrovCommented Feb 5, 2016 at 8:49
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2$\begingroup$ In complex case this question now looks more difficult for me than I initially thought. It is not about matrices, you just have $n$ complex numbers $c_1,\dots,c_n$ such that $\sum c_i=0$, $\sum |c_i|=A$, $\sum |1+c_i|=B$ and you have to describe a locus $(A,B)\subset \mathbb{R}^2$. For example, if $n=2$, it is described by inequalities $\max(A,2)\leqslant B\leqslant \sqrt{A^2+4}$. $\endgroup$– Fedor PetrovCommented Feb 5, 2016 at 9:50
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1 Answer
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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 $\mathrm{tr}|C|$:
- For $\mathrm{tr}|C|\le2(n-2)$, we have $$ 0\le \mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ . $$
- 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 $$\mathrm{tr}|\mathbb I+C|=n\ .$$
- 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 $0\le2s\le \mathrm{tr}|C|$.
- For $\mathrm{tr}|C|\le2(n-2)$, any $0\le 2s\le\mathrm{tr}|C|$ is admissible, giving the general constraint $$ 0\le \mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ . $$
- 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 $$\mathrm{tr}|\mathbb I+C|=n\ .$$
- Finally, for $\mathrm{tr}|C|>2(n-1)$, we have that $0\le s \le n-2$, and thus only have the solution $$ 4-n+\mathrm{tr}|C|\le\mathrm{tr}|\mathbb I+C|\le n +\mathrm{tr}|C|\ . $$
