Prove without using the fact that row rank equals column rank that:
The row rank of a submatrix of $A_{m\times n}$ cannot exceed the row rank of $A$.
My attempt:
I try to use the following result:
Let $x_1,x_2,...,x_k$ be vectors in $F^n$ and let $y_i$ be the vector in $F^{n-1}$ formed by any $n-1$ components of $x_i$ for $i=1,2,...,k$.
If $y_1,y_2,...,y_k$ are linearly independent in $F^{n-1}$ then $x_1,x_2,...,x_k$ are linearly independent in $F^n$.
Let $B=A(i_1,i_2,...i_k|j_1,j_2,...j_l)$ be a submatrix of $A$.
Let row rank of $B=\dim(\mathcal{R}(B))=b$ and row rank of $A=\dim(\mathcal{R}(A))=a$.
Suppose $P=\{B_{{m_1}*},B_{{m_2}*},...,B_{{m_b}*}\}$ where, $m_x\in\{1,2,...,k\}\quad\forall x=1,2,...,b$, is a basis of $\mathcal{R}(B)$.
$\Rightarrow P$ is a linearly independent set.
Using the result stated, we see that $P'=\{A_{i_{m_1}*},A_{i_{m_2}*},...,A_{i_{m_b}*}\}$ is a linearly independent set of some row vectors of $A$ and $|P'|=b$.
Now, the row rank of any matrix is the maximum number of linearly independent row vectors of that matrix and $P'$ is a linearly independent set of row vectors of $A$ with $|P'|=b$
$\Rightarrow a\ge b$
$\Rightarrow$ row rank of $A\ge$ row rank of $B$
Is my approach correct? Is there a better/easier approach?
Note: $\mathcal{R}$ stands for the row space.
Without checking much, you approach is correct, since linear dependence of rows of submatrix $B$ is partly inherited from that of $A$.
There are shorter but not easier proofs, since they just uses same idea.
E.g. (assuming for simplicity that $B$ is obtained by removing first row and first column of $A$, the general case is done easily with same idea, except messier construction.) This is an intuitive proof.
Let $R$ be the matrix representing elementary row operations that maps $B$ to its RREF (of rank $r$). Now let $R'$ be the mattrix obtained from $R$ by adding a top zero row and left-most zero column, and a $1's$ on to left corner.
$R'A$ has a submatrix $RB$ in RREF, now moving this submatrix to left part of $R'A$ by column operation (which doesn't change row rank), hence $R'A$ has at least $r$ leading $1's$. Hence $rank(A)=rank(R'A)\ge r=rank(RB)=rank(B)$
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