Is rank of submatrix less than or equal to rank of matrix?

Question

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?

Answer 1

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 $$ \operatorname{rank} B \leq \min(\operatorname{rank} P, \operatorname{rank} Q, \operatorname{rank} A) \leq \operatorname{rank} A $$

Answer 2

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.

Answer 3

Suppose the matrix is $n\times m$. Let $\alpha\subseteq\{1,2,\ldots,n\}$ and $\beta\subseteq\{1,2,\ldots,m\}$. 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 $A$.