Suppose we have an $n \times n$ square matrix $A$ and we form a submatrix $B$ by taking the first $m$ rows of $A$. Let $A$ have full rank. Is it true, that $B$ has rank $m$ then?
My guess is yes, because if $A$ has $n$ linearly independent rows, then removing some of them does not make the remaining $m$ rows dependent.
Tanks for your help.
Yes, that's right. Vectors are linearly independent if you can't express the zero vector as a linear combination of them. If you can't express the zero vector as a linear combination using all the rows, you also can't do so using $m$ of them.