A a is matrix of dimension $n $ x $ m$ such that $m>n$
(Implies that)$ =>$ there is at least $m-n$ free variables
(Implies that)$=>$There is at least $m-n$ Linear Independent vectors in Solution space of Ax=0
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Now its straight forward to check the first two options,
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A)True.
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Assume there is less than $m-n$ Linearly independent vector solutions to $Ax=0$
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"If those linear independent vectors are solutions to $Ax=0$ then their linear combinations also solution to $Ax=0$."
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(Implies that)$=>$ There is less than $m-n$ free variables.
(Implies that )$=>$ There is atleast $n+1(= m- (m-n-1))$ pivot variables.
(Implies that)$=>$ There is atleast $n+1$ linearly independent vectors in A.
Now just think for moment in a matrix of dimension $n$ x $m$ we'll have almost n pivot variables, but I arrived conclusion that atleast $n+1$ pivot variables which is a contradiction, which means my assumption is incorrect.
So option A is True.
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B) False.
Because its not always the case that every solution in $Ax=0$ is linear combination of $m-n$ linear independent vectors, consider a case where there are less than $n$ Li independent in A then in this case you need more that $m-n$ Linearly independent vectors such that every solution of $Ax=0$ will be linear combination of these vectors.
Ex:-
$A = \left[\begin{array}{ccc} 1 & 2 & 0\\0 & 0 & 0\end{array}\right]$
for this matrix Solution space of Ax=0 will be
$\Biggl\{x_2\left[\begin{array}{c}-2 \\ 1\\ 0\end{array}\right] + x_3\left[\begin{array}{c}0 \\ 0\\ 1\end{array}\right]$ $ \Biggl|$ $ x_2, x_3 \in \mathbb{R} \Biggl\}$,
We can clearly see that by just using $m-n(=1)$ Linear independent vectors we can't generate all solutions in Ax=0.
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Option B would have been True if they mention Number of linear independent vectors in A is equals to n then we can represent every vector in solution space of $Ax=0$ as linear combo of $m-n$ linear independent vectors.
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C)False.
Consider $A = \left[\begin{array}{ccc} 1 & 0 & 2\\0 & 1 & 3\end{array}\right]$
Solution space of $Ax=0$ will be $\Biggl\{x_3\left[\begin{array}{c}-2 \\ -3\\ 1\end{array}\right]$ $\Biggl|$ $ x_3 \in \mathbb{R} \Biggl\}$, Observe carefully
$1) $when $x_3 = 0$ it will be zero solution but our main interest is in NON ZERO solution
$2) $when $x_3 \neq 0$ it will be a non zero solution but every variable is non zero which means there is no non zero solution which has at least $m-n$ zeros.
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It Contradicts Option C.
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D)False.
Consider $A = \left[\begin{array}{ccc} 1 & 0 & 0\\0 & 1 & 0\end{array}\right]$
Solution Space of $Ax=0$ will be, $\Biggl\{x_3\left[\begin{array}{c}0 \\ 0\\ 1\end{array}\right]$ $\Biggl|$ $ x_3 \in \mathbb{R} \Biggl\}$
$1)$ $x_3 =0 $ then it will be zero solution
$2)$ $x_3 \neq 0$ then every solution has exactly one non zero variable but here $n=2$.
It contradicts Option d.
