Question-29

vector space

Consider a set \(V = \{(x, y)\ |\ x, y \in \mathbb{R}\}\) with the usual rule of addition borrowed from \(\mathbb{R}^2\) and scalar multiplication defined as:

\[ c(x, y) = \begin{cases} (0, 0) & c = 0\\ (\frac{cx}{2}, \frac{y}{c}) & c \neq 0 \end{cases} \]

Consider the statements given below:

\(\mathbf{P}\): \(V\) is closed under addition.

\(\mathbf{Q}\): \(V\) has a zero element with respect to addition.

\(\mathbf{R}\): \(1.v=v\) where \(1 \in \mathbb{R}\) and \(v \in V\)

\(\mathbf{S}\) : \(a(v_1 + v_2)=av_1 + av_2\), where \(v_1, v_2 \in V\) and \(a \in \mathbb{R}\)

\(\mathbf{T}\): \((a + b)v = av + bv\), where \(a, b \in \mathbb{R}\) and \(v \in V\)

Choose all correct options

Only \(P\) is true
Only \(Q\) is true
\(P, Q\) and \(S\) are true
Both \(R\) and \(T\) are not true

Answer

Only \(P\) is true
Only \(Q\) is true
\(P, Q\) and \(S\) are true
Both \(R\) and \(T\) are not true

Solution

\(P\) is true as the addition operation is the usual one defined for \(\mathbb{R}^{2}\)
\(Q\) is also true as \((0, 0)\) is an element of \(V\)
\(R\) is not true. Take \(c = 1\) and \((x, y) = (1, 1)\). Then \(1(1, 1) = \left(\cfrac{1}{2}, 1\right) \neq (1, 1)\).
\(S\) is true.
- Let \(v_1 = (x_1, y_1)\) and \(v_2 = (x_2, y_2)\).
- On the LHS we have:
  - \(v_1 + v_2 = (x_1 + x_2, y_1 + y_2)\). Then, \(a(v_1 + v_2) = \left(\cfrac{a}{2}(x_1 + x_2), \cfrac{y_1 + y_2}{a}\right)\).
- On the RHS we have:
  - \(av_1 + av_2 = \left(\cfrac{a}{2} x_1, \cfrac{y_1}{a}\right) + \left( \cfrac{a}{2} x_2, \cfrac{y_2}{a}\right) = \left(\cfrac{a}{2}(x_1 + x_2), \cfrac{y_1 + y_2}{a}\right)\).
- This is for \(a \neq 0\). The proof for \(a = 0\) is quite trivial.
\(T\) is not true. Set \(a = 1, b = 1, v = (1, 1)\). Then \((a + b)v = 2(1, 1) = \left(1, \cfrac{1}{2}\right)\), while \(av + bv = 1(1, 1) + 1(1, 1) = \left(\cfrac{1}{2}, 1\right) + \left(\cfrac{1}{2}, 1\right) = (1, 2)\).