A set \( V \) is given, together with definitions of addition and scalar multiplication. Determine which properties of a vector space are satisfied. (Select all that apply.)
\( V \) is the set of vectors in \( \mathbf{R}^{2} \) with the following definitions of addition and scalar multiplication:
Addition: \( \left[\begin{array}{l}a_{1} \\ b_{1}\end{array}\right]+\left[\begin{array}{l}a_{2} \\ b_{2}\end{array}\right]=\left[\begin{array}{c}0 \\ b_{1}+b_{2}\end{array}\right] \)
Scalar multiplication: \( c\left[\begin{array}{l}a_{1} \\ b_{1}\end{array}\right]=\left[\begin{array}{c}0 \\ c b_{1}\end{array}\right] \).
Property 1: If \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \) are in \( V_{1} \), then so is \( \mathbf{v}_{1}+\mathbf{v}_{2} \).
Property 2: If \( c \) is a real scalar and \( \mathbf{v} \) is in \( V \), then so is \( c \mathbf{v} \).
Property 3: There exists a zero vector \( \mathbf{0} \) in \( V \) such that \( \mathbf{0}+\mathbf{v}=\mathbf{v} \) for all \( \mathbf{v} \) in \( V \).
Property 4: Property 3 holds and for each \( \mathbf{v} \) in \( V \) there exists an additive inverse vector \( -\mathbf{v} \) in \( V \) such that \( \mathbf{v}+(-\mathbf{v})=\mathbf{0} \) for all \( \mathbf{v} \) in \( V \).
Property 5(a): For all \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \) in \( V \), we have \( \mathbf{v}_{1}+\mathbf{v}_{2}=\mathbf{v}_{2}+\mathbf{v}_{1} \).
Property 5(c): For all \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \) in \( V \) and real scalars \( c_{1} \), we have \( c_{1}\left(\mathbf{v}_{1}+\mathbf{v}_{2}\right)=c_{1} \mathbf{v}_{1}+c_{1} \mathbf{v}_{2} \).
Property 5(d): For all \( \mathbf{v}_{1} \) in \( V \) and real scalars \( c_{1} \) and \( c_{2} \), we have \( \left(c_{1}+c_{2}\right) \mathbf{v}_{1}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{1} \).
Property 5(e): For all \( \mathbf{v}_{1} \) in \( V \) and real scalars \( c_{1} \) and \( c_{2} \), we have \( \left(c_{1} c_{2}\right) \mathbf{v}_{1}=c_{1}\left(c_{2} \mathbf{v}_{1}\right) \).
Property 5(f): For all \( \mathbf{v}_{1} \) in \( V \), we have 1 \( \cdot \mathbf{v}_{1}=\mathbf{v}_{1} \).
none of these
