Proving Rank of a matrix is greater than its sub matrix

Question

How can I show that the rank of a matrix is always greater than or equal to the rank of every square matrix thereof.. I mean it is self evident to anyone who knows anything about rank of matrices but how can I prove it?

Answer 1

The rank of a matrix $A$ is the dimension of its column space, that is, the largest number of linearly independent columns of $A$.

So take a submatrix $B$ of $A$, and take a maximal number of linearly independent columns of $B$, let's call then $c_1,c_2,\ldots,c_n$. Now look at the corresponding columns of $d_1,\ldots,d_n$ of $A$, that is, the entries of $c_i$ are entries of $d_i$. Let's show that $d_1,\ldots,d_n$ are linearly independent.

Suppose $\lambda_1 d_1+\cdots\lambda_n d_n=0$. Looking just at the entries of the $d_i$ which correspond to the entries of the $c_i$, we see that $\lambda_1c_1+\cdots+\lambda_n c_n=0$, so linearly independence of $c_i$ implies $\lambda_i=0$. Thus the $d_i$ are linearly independent.

Therefore, the rank of $A$ is at least $n$, which is the rank of $B$.

Answer 2

Any matrix can be transformed to a matrix in upper triangular form. The rank of a matrix in upper-triangular form has a rank equal to the number of non-zero entries on the main diagonal.

Any sub-matrix has a subset of the main diagonal as its main diagonal. It would be there cannot me more non-zero entries on the main diagonal of the sub matrix.