I would like to use Schur's Theorem to prove that the rank of a square matrix $A$ is equal to the size of the largest invertible (square) submatrix of $A$.
The first thing I noticed is that we can assume that the submatrix is in the top left corner of $A$, since we can reorder rows and columns of $A$ without chaning its rank. Let us call reordered matrix $A' = CRA$, where $C$ and $R$ are permutation matrices.
Now I used the Schur's Theorem to write $A' = QTQ^T$, where $Q$ is orthogonal and $T$ upper triangular. We can get submatrix $B_k \in \mathbb{R}^{k \times k}$ as $$ B_k = A[1:k,1:k] = Q[1:k,1:n] T Q^T[1:n,1:k] . $$
The last thing I can see is that $rk(B_k) \le \min(k, r)$, where $r$ is the rank of $A$ and consequently of $T$. For invertible $B_k$ we need $rk(B_k) = k$, which requires $k \le r$, but how do we show that maximum $k$ is $r$?
Are all of the above assumptions correct and how would I go on finishing the proof?