Sum of eigenvalues and singular values How one can prove that for a matrix $A\in \mathbb{C}^{n\times n}$ with eigenvalues $\lambda_i$ and singular values $\sigma_i$, $i=1,\ldots,n$, the following inequality holds:
$$ \sum_{i=1}^n \sigma_i(A) \geq\sum_{i=1}^n \left |\lambda_i(A) \right |$$
 A: *

*Due to Schur decomposition, there exist a unitary matrix $U$ and an upper triangular matrix $T$, such that $A=UTU^{-1}$. Note that $A$ and $T$ share the same eigenvalues and singular values, so we may assume that $A$ is upper triangular.

*Now let $A$ be upper triangular, so the diagonal of $A$ consists of eigenvalues of $A$, and we may assume  the $(i,i)$ entry of $A$ is $\lambda_i(A)$.  For $1\le i\le n$, let $\lambda_i(A)=|\lambda_i(A)|\alpha_i$, where $|\alpha_i|=1$. Denote $D=\mathrm{diag}(\alpha_1,\dots, \alpha_n)$ and let $B=D^{-1}A$. Since $D$ is unitary, $B$ and $A$ share the same singular values. Moreover, by the choice of $D$, we have:
$$\mathrm{tr} B=\sum_{i=1}^n|\lambda_i(A)|.\tag{1}$$

*Let $B=U\Sigma V$ be the singular value decomposition of $B$, i.e. $U$ and $V$ are unitary, and $\Sigma=\mathrm{diag}(\sigma_1(A),\dots, \sigma_n(A))$. Then
$$\mathrm{tr} B=\mathrm{tr}(U\Sigma V)=\mathrm{tr}(\Sigma VU)=\sum_{i=1}^n\sigma_i(A)c_{ii},\tag{2}$$
where $c_{ii}$ is the $(i,i)$ entry of $VU$. Since $VU$ is unitary, $|c_{ii}|\le 1$ for $1\le i\le n$. Then from $(2)$ we know that
$$|\mathrm{tr} B|\le \sum_{i=1}^n\sigma_i(A)|c_{ii}|\le \sum_{i=1}^n\sigma_i(A).\tag{3}$$


The conclusion follows from $(1)$ and $(3)$.

Remark: Denote the collection of $n\times n$ unitary matrix by $U_n$. From the argument above it is easy to see that for every $n\times n$ matrix $A$, we have: (i) there exist $U_0,V_0\in U_n$, such that
$$\sum_{i=1}^n|\lambda_i(A)|=\mathrm{tr} (U_0AV_0);$$
and (ii)
$$\sum_{i=1}^n\sigma_i(A)=\sup_{U,V\in U_n}|\mathrm{tr} (UAV)|.$$
The conclusion is implied by facts (i) and (ii) immediately.
