For a non-null matrix $A_{m \times n}$, show that a submatrix of A formed by $k$ linearly independent rows and $k$ linearly independent columns need not be invertible.
According to me, it should be always invertible. The Gauss Jordan elimination method of finding inverse is premised upon it. How it can be disproved?
Best Answer
Gauss-Jordan elimination only uses invertibility in the special case when $k$ is equal to the rank of $A$. If $k$ is smaller than the rank, the submatrix can fail to be invertible.
For a simple example, let $A=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$. Then the first row of $A$ is linearly independent, as is the first column. However, their intersection is just $\begin{pmatrix} 0\end{pmatrix}$ which is not invertible.