GATE DA LA Practice Problem

Question

Select the correct statement(s) regarding the existence of real matrices:

A) There exists a matrix whose nullspace is $\text{span}\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \\ 0 \\ 0 \end{bmatrix} \right\}$ and whose rowspace contains the vector $\begin{bmatrix} 0\\ 0\\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 0\\ 0\\ 0 \\ 1 \end{bmatrix}$.

B) There exists a $4 \times 3$ matrix with a rowspace dimension of $3$ and a columnspace dimension of $4$.

C) There exists a matrix whose columnspace contains $\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$,$\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}$and whose rowspace contains $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$,$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

D) There exists a matrix whose nullspace consists of all vectors in $\mathbb{R}^3$ satisfying $x_1 +x_2 = 0$, and whose rowspace contains the vector $\begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix}$.

 

Answer 1

a) True. $\text{Row}(A)$$\perp$$N(A)$. Test if the provided vectors are orthogonal:

                         $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} = 0$  and $\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = 0$

Because the dot products between all given basis vectors of the null space and the vectors in the row space evaluate to zero, they are orthogonal.

b) False. A fundamental property of any matrix is that the dimension of its row space must exactly equal the dimension of its column space. Because $3 \neq 4$, such a matrix cannot exist.

c) True. It describes a 3 x 2 matrix. The given column vectors are linearly Independent (rank > 2) and the row vectors are linearly independent (rank > 2). For a 3 x 2 matrix, the maximum possible rank is 2. It is possible to construct a rank-2 matrix that fits these conditions.

d) False. Zero vector is only vector which is present in Row(A) and Null(A) simultaneously.  The null space is defined by $\mathbf{x}_1 + \mathbf{x}_2$= 0. Row space vector $\mathbf{v} = \begin{bmatrix} 2 & -2 & 0 \end{bmatrix}^T$ in this equation (2) + -2(1) + 0(0) = 0 This means $\mathbf{v}$ actually belongs to the null space. The row space and null space are orthogonal complements. A non-zero vector cannot exist in both spaces simultaneously because it would be orthogonal to itself ($\mathbf{v}$.$\mathbf{v}$ = 0).