Question-47

Select all choices that are subspaces of \(\mathbb{R}^{3}\).

Note: \(\mathbb{R}\) denotes the set of real numbers.

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ \mathbf{x} = \alpha \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + \beta \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix},\ \alpha, \beta \in \mathbb{R} \right\}\)

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ \mathbf{x} = \alpha^2 \begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix} + \beta^2 \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix},\ \alpha, \beta \in \mathbb{R} \right\}\)

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ 5x_1 + 2x_3 = 0,\ 4x_1-2x_2+3x_3=0 \right\}\)

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ 5x_1 + 2x_3 + 4 = 0 \right\}\)

Answer

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ \mathbf{x} = \alpha \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + \beta \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix},\ \alpha, \beta \in \mathbb{R} \right\}\)

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ \mathbf{x} = \alpha^2 \begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix} + \beta^2 \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix},\ \alpha, \beta \in \mathbb{R} \right\}\)

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ 5x_1 + 2x_3 = 0,\ 4x_1-2x_2+3x_3=0 \right\}\)

• \(\left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ 5x_1 + 2x_3 + 4 = 0 \right\}\)

Solution

Option-1 \[ \left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ \mathbf{x} = \alpha \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + \beta \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix},\ \alpha, \beta \in \mathbb{R} \right\} \] This can be expressed as the span of the vectors \(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\) and is a subspace. Geometrically, this is a plane in \(\mathbb{R}^{3}\) passing through the origin.

Option-2 \[ \left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ \mathbf{x} = \alpha^2 \begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix} + \beta^2 \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix},\ \alpha, \beta \in \mathbb{R} \right\} \]

This is not a subspace as it is not closed under scalar multiplication. For instance, \(\begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix}\) is an element of the set, but \(\begin{bmatrix}-1 \\ -2 \\ 0\end{bmatrix}\) isn’t.

Option-3 \[ \left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ 5x_1 + 2x_3 = 0,\ 4x_1-2x_2+3x_3=0 \right\} \] This is the intersection of two planes and is hence a subspace of \(\mathbb{R}^{3}\).

Option-4 \[ \left\{ \mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} \in \mathbb{R}^{3}\ :\ 5x_1 + 2x_3 + 4 = 0 \right\} \] This is not a subspace of \(\mathbb{R}^{3}\) as it doesn’t contain the zero vector.