Question

Consider the following subspaces of the real vector space \( \mathbb{R}^3 \): \[ V_1 = \text{span} \left\{ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right\}, \quad V_2 = \text{span} \left\{ \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right\}, \quad V_3 = \text{span} \left\{ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right\}, \quad V_4 = \text{span} \left\{ \begin{pmatrix} 1 \\ 3 \\ 6 \end{pmatrix} \right\}, \quad V_5 = \text{span} \left\{ \begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} \right\}. \] Then, which of the following is/are TRUE?

• A. \( V_1 \cup V_2 \) is a subspace of \( \mathbb{R}^3 \).
• B. \( V_1 \cup V_3 \) is a subspace of \( \mathbb{R}^3 \).
• C. \( V_1 \cup V_4 \) is a subspace of \( \mathbb{R}^3 \).
• D. \( V_1 \cup V_5 \) is a subspace of \( \mathbb{R}^3 \).

Hint:

The union of two subspaces is not necessarily a subspace unless one of the subspaces is contained within the other. Always check closure under addition and scalar multiplication.

Answer 1

The Correct Option is C

Step 1: Check if the union of subspaces forms a subspace.
For a union of subspaces to be a subspace, it must satisfy the properties of closure under addition and scalar multiplication.
We check the options one by one:

- \( V_1 \cup V_2 \): The union of two subspaces is not necessarily a subspace unless one is contained within the other.
Here, the set \( V_2 \) is a one-dimensional subspace, and \( V_1 \) is a two-dimensional subspace.
Since they are not scalar multiples of each other, their union is not closed under addition or scalar multiplication.
Therefore, \( V_1 \cup V_2 \) is not a subspace.

- \( V_1 \cup V_3 \): The union of \( V_1 \) and \( V_3 \) is not closed under addition.
Specifically, the sum of two vectors from these subspaces may not lie in the union.
Therefore, \( V_1 \cup V_3 \) is not a subspace.

- \( V_1 \cup V_4 \): The subspaces \( V_1 \) and \( V_4 \) are both spanned by vectors in \( \mathbb{R}^3 \).
The span of \( V_1 \) and \( V_4 \) together forms a two-dimensional subspace, which is closed under addition and scalar multiplication.
Hence, \( V_1 \cup V_4 \) is a subspace of \( \mathbb{R}^3 \).

- \( V_1 \cup V_5 \): The union of these two subspaces is not closed under addition,
as elements from \( V_1 \) and \( V_5 \) may not add up to an element in either subspace.
Therefore, \( V_1 \cup V_5 \) is not a subspace.

Final Answer:
\[ \boxed{V_1 \cup V_4 \text{ is a subspace of } \mathbb{R}^3.} \]