Exercise Sheet 1 Solutions

Contents

Exercise Sheet 1 Solutions#

1.#

(a)#

Take any v1=(a,b) and v2=(c,d) in V; then b=3a+1 and d=3c+1.
Their sum is

v1+v2=(a+c,b+d)=(a+c,3a+1+3c+1)=(a+c,3(a+c)+2),

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)V, violating axiom V1.)

(b)#

All axioms except distributivity over scalar addition fail:

Take v=(a,b) and scalars α,βR.

(α+β)v=((α+β)a,b),αv+βv=(αa,b)+(βa,b)=((α+β)a,2b).

Unless b=0, the second component differs, so
(α+β)vαv+βv.
Therefore V is not a vector space.

2.#

(a)#

Zero vector: (0,0) satisfies 0=20.
Closure (addition): if y1=2x1 and y2=2x2, then

y1+y2=2(x1+x2).

Closure (scalar mult.): for αR,

α(x,y)=(αx,2αx).

All three conditions hold ⇒ W is a subspace.

(b)#

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

α(x,y)=(αx,αy),

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

3.#

For x=(a,b), y=(c,d) and scalars α,β:

T(αx+βy)=((αa+βc)2,αb+βd),
αT(x)+βT(y)=(α2a2+β2c2,αb+βd).

The first components differ unless ac=0 or αβ=0.
Hence T violates additivity/homogeneity ⇒ not linear.