Question 49

For any $n \times m$ matrix $A$ with $m > n$, the rank $r = \text{rank}(A)$ satisfies $r \le n$. By the Rank-Nullity Theorem, the dimension of the solution space (null space) of $Ax = 0$ is:

• (A) is TRUE: Since the solution subspace has dimension at least $m - n$, any basis of the null space contains at least $m - n$ linearly independent solutions.
• (B) is FALSE: If $r < n$ (for example, $A = 0$), the null space has dimension $m - r > m - n$, so $m - n$ vectors cannot span the entire solution space.
• (C) is FALSE: Putting $A$ in reduced row echelon form gives $r \le n$ pivot variables and $m - r \ge m - n$ free variables; setting one free variable to $1$ and the remaining $(m - r - 1)$ free variables to $0$ only guarantees $m - n - 1$ zero entries. For example, if $n = 1, m = 2$ and $A = \begin{bmatrix} 1 & -1 \end{bmatrix}$, every non-zero solution to $x_1 - x_2 = 0$ has $x_1 = x_2 \neq 0$ ($0$ variables equal to $0$, whereas $m - n = 1$).
• (D) is FALSE: If $n = 2, m = 3$ and $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$, every solution to $Ax = 0$ has $x_1 = 0, x_2 = 0$, so at most $1 < n = 2$ variable can be non-zero.