Inequality between the rank of matrix and the $\ell_1$-norm Let $B=[b_{ij}]\in M_n(\mathbb C)$ and $b_j$ denote the $j$th column of $B$.

Problem: Prove that $$\text{rank} B\geq\sum_{b_j \neq 0} \frac{|b_{jj}|}{\| b_j\|_1}$$
where $\|b_j\|_1 = \sum_{j=1}^n |b_j|.$

I don't have any idea or approach on how to show this inequality. Any help and hints would be much appreciated.

Sketch of proof: Find an invertible, diagonal matrix $D$ such that all columns of the matrix $M = BD$ have $1$-norm $0$ or $1$ and the diagonal entries of $M$ are all non-negative. Observe that the rank of $M$ is equal to the rank of $B$.
On the other hand, note that $\|M\|_1 \leq 1$, which implies that all eigenvalues of $M$ have magnitude at most $1$. Conclude that
$$
|\operatorname{tr}(M)| \leq \operatorname{rank}(M).
$$
