I'm quite confused when reading Horn and Johnson's Matrix analysis (second edition) p.20 , 0.7.6 Rank in a partitioned matrix and rank-principal matrices .
Denote $M_{m,n}(F)$ as the set of all $m\times n $ matrices over field $F$ (arbitrary) , for square matrices use abbreviation . Partition $A \in M_n(F)$ as $$ A = \left(\begin{array}{rr} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right) ,\quad A_{11} \in M_r(F) , A_{22} \in M_{n-r}(F) $$ The book attempts to prove (actually it's $\iff$ but I don't think it's relevant here)
$rk(A)=rk[A_{11}\;A_{12} ] = rk\begin{bmatrix}A_{11}\\ A_{21}\end{bmatrix} \implies A_{11}$ is nonsingular .
as follow
Suppose $A_{11}$ is singular , then $rk(A_{11})=k<r $ . There are nonsingular $S,T\in M_r(F)$ such that $$ SA_{11}T = \left(\begin{array}{rr} I_{k} & 0 \\ 0 & 0_{r-k} \end{array}\right) $$ Therefore, $$ \hat{A} = \left(\begin{array}{rr} S & 0 \\ 0 & I_{n-r} \end{array}\right) A \left(\begin{array}{rr} T & 0 \\ 0 & I_{n-r} \end{array}\right) = \left(\begin{array}{rr} SA_{11}T & SA_{12} \\ A_{21}T & A_{22} \end{array}\right) $$ has rank $r$ , as do its block row and column . Because the $r^{th}$ row of the first block column of $\hat{A} $ is zero , there must be some column in $SA_{12}$ whose $r^{th}$ entry is not zero, which means that $\hat{A} $ has at least $r + 1$ independent columns. This contradicts $rank \hat{A} = rank A = r$ ,so $A_{11}$ must be nonsingular. $\square$
Question : Why $rank \hat{A} = rank A = r ? $ I only know $rank \hat{A} = rank A$ since multiplication with nonsingular matrices won't change rank . I don't recall making any assumption on $A_{12}\in M_{r,n-r}(F),A_{21}\in M_{n-r,r}(F)$ and $A_{22}$ , so let all of them be $0$ , then $rk(\hat{A}) = k < r $ .
Nonetheless , if assume $rank \hat{A} = rank A = r $ , then the rest of the proof follows . I couldn't find such assumption . Am I missing something ?