When $m=n$, we get unique solution that is trivial and hence linearly dependent. So, we get $0$ linearly independent solution (Here, $n-m=0$).
When $m<n$, we also get non-trivial solution along with the trivial solution. We discard the trivial solution because we want only linearly independent solutions. Since we have $m$ linearly independent rows, the rank of $m \times n$ matrix is $m$. Hence $n-m$ linearly independent solutions
The number of linearly independent rows is equal to number of linearly independent columns. Thus, number of linearly independent rows cannot exceed the number of columns. So, $m>n$ is an impossible case.
The correct option is $(C)$ $n-m$.
Why are the colums of
$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$
linearly dependent?
Because there exists non-zero $x$ such that
$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix} x = 0$
i.e.
$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}\begin{bmatrix}
0\\0\\0\\1\end{bmatrix} = \begin{bmatrix} 0\\0\\0\end{bmatrix}$
How do you prove that any $3\times4$ matrix has linearly dependent columns?
Suppose the columns of your matrix are $\mathbf v_1,\mathbf v_2,\mathbf v_3,\mathbf v_4.$ And suppose that $\mathbf v_1,\mathbf v_2,\mathbf v_3$ are linearly independent. Then we want to show that there exists and $a,b,c$ such that $a\mathbf v_1 + b\mathbf v_2 + c\mathbf v_3 = \mathbf v_4$
How to do that? It might help to show that there exists $a_1,b_1,c_1$ such that:
$a_1\mathbf v_1 + b_1\mathbf v_2 + c_1\mathbf v_3 = \begin{bmatrix} 1\\0\\0\end{bmatrix}$
and similarly there is $a_2,b_2, c_2$ and $a_3, b_3, c_3$ such that
$a_2\mathbf v_1 + b_2\mathbf v_2 + c_2\mathbf v_3 = \begin{bmatrix} 0\\1\\0\end{bmatrix}$
and
$a_3\mathbf v_1 + b_3\mathbf v_2 + c_3\mathbf v_3 = \begin{bmatrix} 0\\0\\1\end{bmatrix}$
And certainly $\mathbf v_4$ can be composed as a combintation of $\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0\end{bmatrix},\begin{bmatrix} 0\\0\\1\end{bmatrix}$
Best Answer
Yes, you can rather work for the contrapositive statement. Suppose columns of $A$ are linearly dependent then $rank(A)<n$, where $n$ is number of columns in $A$ or equivalently number of variables in the system of equations furnished by $AX=b_1$. Now $rank(A)<n$ implies either infinitely many solutions or no solution i.e. non-uniqueness in either case.