Question

Let \[ M = \begin{pmatrix} 1 & -1 & 1 \\ 1 & -1 & -1 \end{pmatrix}. \] If \[ V = \{ (x, y, 0) \in \mathbb{R}^3 : M \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \}, \] \[ W = \{ (x, y, z) \in \mathbb{R}^3 : M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \}, \] then

• A. the dimension of \( V \) equals 2
• B. the dimension of \( W \) equals 2
• C. the dimension of \( V \) equals 1
• D. \( V \cap W = \{(0,0,0)\} \)

Hint:

To find the dimension of the null space, solve the equation \( M \mathbf{x} = 0 \) and count the number of free variables in the solution.

Answer 1

The Correct Option is C

Given: $M = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -1 & -1 \end{bmatrix}$

Finding $V$: Points $(x,y,0)$ where $M\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$$M\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = \begin{bmatrix} x - y \\ x - y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

This gives: $x - y = 0$, so $x = y$.

Therefore: $V = {(x, x, 0) : x \in \mathbb{R}} = {x(1, 1, 0) : x \in \mathbb{R}}$

Dimension of $V$ = 1

Finding $W$: Points $(x,y,z)$ where $M\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$$M\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x - y + z \\ x - y - z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

From the equations:

• $x - y + z = 0$ ... (1)
• $x - y - z = 0$ ... (2)

Adding: $2(x - y) = 0 \Rightarrow x = y$

Subtracting: $2z = 0 \Rightarrow z = 0$

Therefore: $W = {(x, x, 0) : x \in \mathbb{R}} = V$

Dimension of $W$ = 1

Verify options:

(A) dim($V$) = 2? FALSE 

(B) dim($W$) = 2? FALSE 

(C) dim($V$) = 1? TRUE 

(D) $V \cap W = {(0,0,0)}$? FALSE (since $V = W$) 

Answer: (C)