Question

If A is a $3\times 3$ matrix non-zero matrix, then prove that the rank of the matrix is
[i] 1 if Every $2\times 2$ submatrix of A is singular
[ii] 2 if there exists a $2\times 2$ submatrix of A which is non-singular and A is singular
[iii] 3 if A is non-singular.

Answer 1

Hint: Use the fact that columns and the rows of a non-singular matrix are linearly independent. Hence argue that if an nxn matrix has $m\times m$submatrix which is non-singular, then $rank\left( A \right)\ge m$. Use the fact that if an nxn matrix is singular, then the matrix cannot be a full rank matrix. Hence prove the statements (i), (ii) and (iii).

Complete step by step answer:

Property: If an $n\times n$ matrix A has a non-zero determinant, then the following statement are true:
[a] At least one $\left( n-1 \right)\times \left( n-1 \right)$ submatrix of A is non-singular (Converse does not hold)
[b] The rows of the matrix are Linearly independent (Converse of this also holds)
[c] The columns of the matrix are Linearly independent (Converse of this also holds)
[d] A does not have 0 as its eigenvalue (Converse also holds).
[e] $rank\left( A \right)=n$(Converse also holds)
We use the above properties to prove the statements of the question
Statement [i]:
Since every $2\times 2$ submatrix of A is singular, we have A is singular (From property [a])
Hence, we have $rank\left( A \right)\le 2$.
Now if rank(A) = 2, we have two rows of the matrix are Linearly independent. Let the rows be ${{c}_{1}}=\left[ {{c}_{11}},{{c}_{12}},{{c}_{13}} \right]$ and ${{c}_{2}}=\left[ {{c}_{21}},{{c}_{22}},{{c}_{23}} \right]$. Since the rows are linearly independent, there exists a non-singular submatrix of the matrix $\left[ \begin{matrix}
   {{c}_{11}} & {{c}_{12}} & {{c}_{13}} \\
   {{c}_{21}} & {{c}_{22}} & {{c}_{23}} \\
\end{matrix} \right]$. Since every $2\times 2$ submatrix of $\left[ \begin{matrix}
   {{c}_{11}} & {{c}_{12}} & {{c}_{13}} \\
   {{c}_{21}} & {{c}_{22}} & {{c}_{23}} \\
\end{matrix} \right]$ is a submatrix of the original matrix A, we have there exists a non-singular $2\times 2$ submatrix of A which is not possible.
Hence, we have $rank\left( A \right)\ne 2$
Hence, we have
$rank\left( A \right)\le 1$
Also, since A is non-zero matrix, we have $rank\left( A \right)\ge 1$
Hence, we have $rank\left( A \right)=1$
[ii] Since A is singular, we have $rank\left( A \right)\le 3-1=2$(Since if rank(A) = 3, then A has to be non-singular by property [e]).
Also, since there exists a non-singular $2\times 2$ submatrix of A, there exists two linearly independent columns of A. Hence, we have $rank\left( A \right)\ge 2$
Hence, we have $rank\left( A \right)=2$
[iii] Although this is simply the property [e] , we prove the statement as follows.
We know that A is non-singular. Two cases arise.
Case I: $rank\left( A \right)=3$. In this case, we have nothing to do.
Case II: $rank\left( A \right)\le 2$
Since A is not a full rank matrix, there exists 0 Row in its row reduced form. Let the row reduced form matrix of A be R.
Hence, we have $R=EA$, where E is the elementary matrix of A. Since E is invertible, we have E is non-singular.
Hence, we have $\det \left( E \right)\ne 0$
Also. since the last row of R is 0, we have det(R) = 0
Now, we know that $\det \left( EA \right)=\det \left( E \right)\det \left( A \right)=\det \left( R \right)=0$
Since $\det \left( E \right)\ne 0$, we have $\det \left( A \right)=0$.
Hence, A is singular. This contradicts the fact that A is non-singular.
Hence, we have rank(A) = 3
Q.E.D

Note:
[1] These properties should be remembered. The rank of a matrix is an important determining factor in solving linear equation using matrices and holds high importance in linear algebra as it is one of the terms in one of the most theorems in Linear Algebra, i.e. The Rank-Nullity theorem