Rank of Jordan block

Question

Let matrix $A=(a_{ij})_{n\times n } $ be such that $a_{ii}=\lambda, 2\leq i \leq n $ and $a_{ii+1}=1, 1\leq i \leq n-1 $ and $a_{ij}=0$ if $j\not\in \{1,i, i+1\}$ so basically A is nearly a Jordan block of degree n with eigenvalue $\lambda $ except for the first column. Is it possible to show that $$\text{rank} (A-\lambda I_n)=n-1 $$?