Rank of the following matrix

Question

Given that, $\mathrm{rank}\begin{bmatrix}\mu\mathbf{I}+\mathbf{A}&\mathbf{B}&\mathbf{E}\\ \mathbf{C}&\mu\mathbf{I}+\mathbf{D}&\mathbf{F}\end{bmatrix}=\text{full row-rank}=n$ for every $\mu\in\mathbb{C}$. Can we claim that, $\mathrm{rank}\begin{bmatrix}\mu\mathbf{I}+\mathbf{A}&\mathbf{B}&\mathbf{E}&\vdots&\star\\ \mathbf{C}&\mu\mathbf{I}+\mathbf{D}&\mathbf{F}&\vdots&\star\\ \hdashline \mathbf{O}&\mu\mathbf{I}&\mathbf{O}&\vdots&\mathbf{I}_k\end{bmatrix}=n+k$ for every $\mu\in\mathbb{C}?$ I know that if we could reduce this augmented matrix into upper triangular (which happens when $\mu=0$), then certainly this claim is correct. But what about the general case?

Answer 1

No. E.g. $\operatorname{rank}\pmatrix{\mu&1&1\\ 1&\mu-1&1}\equiv2=\operatorname{rank}\left(\begin{array}{ccc|c} 1&1&1&1\\ 1&1-1&1&0\\ \hline 0&1&0&1 \end{array}\right)$.