The set $V$ is defined as:

$V = \left\{ \begin{bmatrix} X \\ Y \\ 0 \end{bmatrix} : X, Y \in \mathbb{R} \right\}$

This represents a subspace of $\mathbb{R}^3$ where the third component is always zero, and $X$ and $Y$ are real numbers.

Step 1: Geometric Interpretation of $V$

This set describes the $XY$-plane in $\mathbb{R}^3$, since the third component is always 0. Hence, this subspace consists of all vectors that lie in the plane spanned by the $X$-axis and the $Y$-axis within the $XY$-plane of $\mathbb{R}^3$.

Step 2: Basis for $V$

To find a basis for $V$, we need to find a set of vectors that:

1. Span $V$: This means we need vectors such that any vector in $V$ can be written as a linear combination of these basis vectors.
2. Are linearly independent: The vectors in the basis must not be linear combinations of each other.

Step 3: Expressing Vectors in $V$

Any vector in $V$ can be written as:

$\begin{bmatrix} X \\ Y \\ 0 \end{bmatrix}$

This can be decomposed as:

$X \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + Y \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$

This shows that the set of vectors:

$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$

spans $V$, since any vector in $V$ can be expressed as a linear combination of these two vectors. These vectors are also linearly independent, as no scalar multiple of one vector can result in the other.

Step 4: Conclusion

Therefore, a basis for $V$ is:

$\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\}$

Additional Questions You Could Explore:

1. How can we verify that a set of vectors is linearly independent?
2. How would the basis change if the third component were nonzero in the definition of $V$?
3. What is the dimension of the subspace $V$?
4. Can we find an alternative set of basis vectors for $V$?
5. How do you express a vector in $V$ in terms of these basis vectors?

Tip: The number of vectors in a basis for a subspace gives you its dimension! In this case, the dimension of $V$ is 2.