Vector Space

Vector Space

1. Give two examples of vector space and show that they fulfill all the axioms. a) V is the set of all 2 × 2 matrices abcd

V is a vector space Solution : u=a1b1c1d1,v= a2b2c2d2,w= a3b3c3d3, c=g,d=f

i. V is closed under addition u+v ∈ V a1b1c1d1+a2b2c2d2 =a1+a2b1+b2c1+c2d1+d2

V

ii. Commutative on addition u+v=v+u u+v=a1b1c1d1+a2b2c2d2 =a1+a2b1+b2c1+c2d1+d2 =a2+a1b2+b1c2+c1d2+d1 =a2b2c2d2+a1b1c1d1 =u+v

iii. Associative on addition u+v+w=u+v+w u+v+w=a1b1c1d1+a2b2c2d2+a3b3c3d3 =a1+a2b1+b2c1+c2d1+d2+a3b3c3d3 =a3+a1+a2b3+b1+b2c3+c1+c2d3+d1+d2 =a1+a2+a3b1+b2+b3c1+c2+c3d1+d2+d3 =a1b1c1d1+a2+a3b2+b3c2+c3d2+d3 1

=a1b1c1d1+a2b2c2d2+a3b3c3d3 =u+(v+w)

iv. Vector identity in addition º=0000 a1b1c1d1+0000

=a1+0b1+0c1+0d1+0 =a1b1c1d1 v. Vector inverse in addition v=a2b2c2d2,-v=-a2-b2-c2-d2 v+-v=a2b2c2d2+-a2-b2-c2-d2 =a2+-a2b2+-b2c2+-c2d2+-d2 =0000 =0v vi. V is closed under scalar multiplication cu∈V cu=ga1b1c1d1 =ga1gb1gc1gd1∈V

vii. Left distributive law cu+v=cu+cv cu+v=ga1b1c1d1+a2b2c2d2 =ga1+a2b1+b2c1+c2d1+d2 =ga1+a2gb1+b2gc1+c2gd1+d2 =ga1+ga2gb1+gb2gc1+gc2gd1+gd2 =ga1gb1gc1gd1+ga2gb2gc2gd2 =ga1b1c1d1+ga2b2c2d2 =cu+cv

viii.Right distributive law c+du=cu+du c+du=g+fa1b1c1d1 =g+fa1g+fb1g+fc1g+fd1 2

=ga1+fa1gb1+fb1gc1+fc1gd1+fd1 =ga1gb1gc1gd1+fa1fb1fc1fd1 =ga1b1c1d1+fa1b1c1d1 =cu+du

ix. Associative in scalar multiplication c( du) = (cd )u    c( du) = g  f  a1 b1    c1 d 1  f f b1   = g  a1 f f d 1   c1  gf gf b1   =  a1 gf gf d 1   c1  gf gf b1   =  a1 gf gf d 1   c1

[ ] [ ]

 = gf  a1  c1 = (cd )u

[ ] [ ]

b  d  1

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x. 1.u=u 1.u=1a1b1c1d1 =a1b1c1d1 =u ∴Thus,V is a vector space

b) The set C of all complex numbers,with the usual operations of addition of complex

numbers and multiplication by real numbers,is a vector space(over the real numbers) Solution : Recall that C is the set of all elements of the form z=a+bi,where a and b are real numbers. Two complex numbers a + bi and c + diare equal if and only if a = c and b = d. Addition is defined by 3

a+bi+c+di=a+c+b+di

and scalar multiplication by ka+bi=ka+kbi. Since each of these is an element of C,axioms and of the definition of vector space are satisfied. . Commutative on addition For

z = a + b i and z = a + b i ( z + z ) = (a + b i) + (a + b i) = ( a + a ) + ( b + b )i = ( a + a ) + ( b + b )i = (a + b i) + (a + b i) =z +z 1

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. Associative on addition For z1 = a1 + b1 i ,

z = a + b i , and z = a + b i ( z + z ) + z = [( a + b i ) + ( a + b i )] + ( a + b i ) = [ ( a + a ) + (b + b )i ] + ( a + b i ) = [ ( a + a ) + a ] + [(b + b ) + b ]i = [ a + ( a + a ) ] + [b + (b + b ) ]i = ( a + b i ) + [ ( a + a ) + ( b + b )i ] = ( a + b i ) + [( a + b i ) + ( a + b i )] = z + (z + z ) 1

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. Vector identity in addition 0 + 0i serves as the zero element,since ( a + bi) + ( 0 + 0i ) = ( a + 0) + ( b + 0) i = a + bi

vi. Vector inverse in addition 4

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For z = a + bi − z = − a − bi since ( a + bi) + ( − a − bi) = 0 + 0i

vii. Associative in scalar multiplication

For z = a + bi and real numbers c and d ( cd )( a + bi) = ( cd ) a + ( cd ) bi = c( da) + c( db) i = c( da + dbi) = c[ d ( a + bi) ]

viii. Left distributive law

z = a + b i , z = a + b i and real number c c ( z + z ) = c [( a + b i ) + ( a + b i )] = c[( a + a ) + (b + b )i ] = c ( a + a ) + c (b + b )i = (c a + c a ) + ( c b + c b )i = (c a + c b i ) + ( c a + c b i ) = c(a + b i ) + c( a + b i ) = c z +c z For 1

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. Right distributive law

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For z = a + bi and real numbers c and d , ( c + d ) z = ( c + d )( a + bi) = ( c + d ) a + ( c + d ) bi = ( ca + da) + ( cb + db) i = ( ca + cbi ) + ( da + dbi) = c ( a + bi) + d ( a + bi ) = cz + dz

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For z = a + bi 1z = 1( a + bi)

= (1a ) + (1b ) i = a + bi =z

∴Thus, C is a vector space

2. Give three examples of vectors that are not vector space. Give counterexamples to show that they are not a vector space. a) Let S be the set of ordered pairs in R2 with + defined by Type equation here.

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