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:
$$ \sum_{i=1}^n \sigma_i(A) \geq\sum_{i=1}^n \left |\lambda_i(A) \right |$$
Sum of Eigenvalues and Singular Values in Linear Algebra
eigenvalues-eigenvectorsinequalitylinear algebramatricessingular values
Best Answer
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.