Comparing the ranks of leading princial minors of a square symmetric matrix

Question

Let $A$ be a $n$ x $n$ real symmetric matrix and let $A_k$ denotes the $k$-th order leading principal minor matrix of $A$. Prove that for $0 \leq k \leq n-1$: $$Rank(A_{k+1})\leq Rank(A_k)+2$$

Comments:

This is very evident by looking at the row echelon form of the matrix. I wonder if there is any other concrete way of proving this.

Thanks in advance!

Answer 1

Let $$ A_{k+1}=\begin{bmatrix}A_k & b\\c^T & d\end{bmatrix}. $$ Adding one row $c^T$ to $A_k$ will increase the rank of $A_k$ by at most $1$ (if $c^T$ is linear independent from the rows of $A_k$), hence $$ \operatorname{rank}\begin{bmatrix}A_k\\c^T\end{bmatrix}\le\operatorname{rank}A_k+1. $$ Now repeat the argument for adding one column $\begin{bmatrix}b\\d\end{bmatrix}$.