Exercise Sheet 1 Solutions

1.

(a)

Take any \(v_1=(a,b)\) and \(v_2=(c,d)\) in \(V\); then \(b=3a+1\) and \(d=3c+1\).
Their sum is

\[ v_1+v_2=(a+c,\;b+d)=(a+c,\;3a+1+3c+1)=\bigl(a+c,\;3(a+c)+2\bigr), \]

which does not satisfy \(b+d=3(a+c)+1\). Hence \(V\) is not closed under addition ⇒ not a vector space.
(Equivalently, the additive identity \((0,0)\notin V\), violating axiom V1.)

(b)

All axioms except distributivity over scalar addition fail:

Take \(v=(a,b)\) and scalars \(\alpha,\beta\in\mathbb R\).

\[ (\alpha+\beta)\,v=((\alpha+\beta)a,\;b), \quad \alpha v+\beta v=(\alpha a,\;b)+(\beta a,\;b)=((\alpha+\beta)a,\;2b). \]

Unless \(b=0\), the second component differs, so
\((\alpha+\beta)v\neq\alpha v+\beta v\).
Therefore \(V\) is not a vector space.

2.

(a)

Zero vector: \((0,0)\) satisfies \(0=2\cdot0\).
Closure (addition): if \(y_1=2x_1\) and \(y_2=2x_2\), then

\[ y_1+y_2 = 2(x_1+x_2). \]

Closure (scalar mult.): for \(\alpha\in\mathbb R\),

\[ \alpha(x,y)=(\alpha x,\;2\alpha x). \]

All three conditions hold ⇒ \(W\) is a subspace.

(b)

Pick \((x,y)\in W\) with \(x>0\) and any negative scalar \(\alpha<0\).
Then

\[ \alpha(x,y)=(\alpha x,\;\alpha y), \]

and \(\alpha x<0\). Thus \(\alpha(x,y)\notin W\).
Not closed under scalar multiplication ⇒ not a subspace.

3.

For \(x=(a,b)\), \(y=(c,d)\) and scalars \(\alpha,\beta\):

\[ T(\alpha x+\beta y)=\bigl((\alpha a+\beta c)^{2},\;\alpha b+\beta d\bigr), \]

\[ \alpha T(x)+\beta T(y)=\bigl(\alpha^{2}a^{2}+\beta^{2}c^{2},\;\alpha b+\beta d\bigr). \]

The first components differ unless \(a c=0\) or \(\alpha\beta=0\).
Hence \(T\) violates additivity/homogeneity ⇒ not linear.