Is rank of submatrix less than or equal to rank of matrix? OK, so I realize this might be a stupid question but an answer can certainly help me in my matrix theory class, I need to know if in general the rank of a submatrix is less than or equal to the rank of the larger matrix? Is it true in general?
 A: Taking submatrix may be expressed using matrix multiplication.
Assume that matrix $B$ is a submatrix of $A$:
$$
B_{i,j} = A_{r_i, c_j}.
$$
Consider matrices $P$ and $Q$ defined as
$$
P_{i, k} = \begin{cases}
1, &k = r_i\\
0, &k \neq r_i
\end{cases}, \qquad
Q_{j, m} = \begin{cases}
1, &m = c_j\\
0, &m \neq c_j
\end{cases}.
$$
The product $PAQ^\top$ is exactly $B$:
$$
(PAQ^\top)_{i,j} = \sum_{k,m} P_{i,k} A_{k,m} Q_{j, m} = A_{r_i, c_j} = B_{i,j}
$$
since
$$
\operatorname{rank} XY \leq \min(\operatorname{rank} X, \operatorname{rank} Y),
$$
we get
$$
\operatorname{rank} B \leq \min(\operatorname{rank} P, \operatorname{rank} Q, \operatorname{rank} A) \leq \operatorname{rank} A
$$
A: The rank of a submatrix is never larger than the rank of the matrix, but it may be equal. 
Here are two simple examples. 
If a $m \times n$ rectangular matrix has full rank $m$, its rank equals the rank of a $m \times m$ submatrix.
If a $m \times m$ square matrix has not full rank, then its rank equals the rank of a submatrix.
A: Suppose the matrix is $n\times m$. Let $\alpha\subseteq\{1,2,\ldots,n\}$ and $\beta\subseteq\{1,2,\ldots,n\}$. Since the rank of a matrix is equal to both its row rank and its column rank,


*

*the rank of $A[\alpha|\beta]$ is less than or equal to the rank of $A\left[\alpha\right|\left.\{1,2,\ldots,m\}\right]$, because the column space of the former is contained in the column space of the latter;

*the rank of $A\left[\alpha\right|\left.\{1,2,\ldots,m\}\right]$ is less than or equal to the rank of $A$, because the row space of the former is contained in the row space of the latter.


Consequently, The rank of $A[\alpha|\beta]$ is always less than or equal to the rank of $A$.
