UNIT - I LINEAR ALGEBRA. , such that αν V satisfying following condition

UNIT - I LINEAR ALGEBRA Definition Vector Space : A non-empty set V is said to be vector space over the field F. If V is an abelian group under addition and if for every α, β F, ν, ν 2 V, such that αν V satisfying following condition (i) α F, v, v α v+ v2 = αv + α v2; (ii) ( + ) v = v + v V α β α β ; α, β F, v V (iii) v = ( v ) α, β F, v V αβ α β ; (iv) F v = v. Example Let V = F [x] over F it is a vector space usual addition and multiplication of polynomial. Subspace : subspace. Let W φ V and α, β F, w, w2 W wth αw+ β w2 W then we called W is Example : W is the collection of all polynomial with degree less than n is subspace of F[ x ]. Homomorphism in vector space If U and V are vector space over F then the mapping T : U V is said to be a homomorphism. If (i) T ( u + u ) = T( u ) + T( u ) (ii) T ( u ) = T( u ) α α u, u U and α F

Lemma : If V is a vector over F then ) α 0= 0 2) 0 v = 0 3) α( v) = ( αv) 4) If v 0 and α v = 0 α = 0, α Fv, V. Lemma : If V is a vector space over F and if W is a subspace of V then V/W is a vector space over F, where for v + W, v 2 + W V/ W (i) v + W + v + W = v + v + W (ii) α ( α ) v + W = v + W Theorem : If T is homomorphism of U onto V with Kernal W then V is isomorphic to V/W, conversely, if U is a vector space and W is subspace of U then there is a homomorphism of U onto U/W. U v T onto T v V η v+ W φ U/W 2

Definition Internal Direct Sum Let V be a vector space over F and let U, U 2,..., U n be subspace of V, V is said to be the internal direct sum of U, U 2,..., v= u + u2 +... + u n where ui Ui. U. If every element v V can be written in one and only one as n Definition External Direct Product Any finite number of vector spaces over F, V, V 2,..., V n. Consider the set V of all order n tuples ( v, v 2,..., v n) where ) Let v= u+ w v i V i, V is called external direct sum V, V 2,..., V n. ( u, u,..., u ) ( v, v,..., v ) = + m (,,..., ) = u + v u + v u + v, uw, V 2 2) αv α( u, u,..., u ) Theorem = m ( αu, αu,..., αu ) = m m m m If V is the internal direct sum of U, U2,..., Unthen V is isomorphic to the external direct sum of U, U2,..., U n. v = u + u 2 +... + u n u, u 2,..., u n Linear Independent and Basis Linear Combination If V be a vector space over F and v, v 2,..., v n V then any element of the form, α v + α2v2 +... + α n v n where α i F is a linear combination over F of v, v2,..., v n. Linear Span If S be a non-empty subset of vector space V then L (S) the Linear Span of S is the set of all linear combinations of elements of S. 3

Lemma : L (S) is subspace of V. If S, T are subsets of V then ) S T LS LT 2) L L( S) = L S 3) L( S T) = L( S) + L( T) Finite Dimensional V The vector space V is said to be finite dimensional if there is a finite subset S in V such that = LS. Linear Dependent Set : If V is vector space and if v, v2,..., v n are in V. We say that they are linearly dependent over F if there exist elements α, α2,..., α n in F not all of them zero such that, α v + α2v2 +... + αnvn= 0 Linearly Independent Set : If V is vector space and if v, v2,..., v n are in V we say that they are linearly independent in F all are zero such that α v + α2v2 +... + α v = 0 n n Lemma If v, v,..., v V are linearly independent then every element in their linear span have a n unique representation in the form α v + α2v2 +... + α n v n with α i ' s F. Theorem If v v 2,,..., v n are in V then either they are linearly independent or some v k is linear combination of preceeding once v, v 2,..., v k. 4

Carollary : If v, v2,..., v n in V have W as a linear span and if v, v 2,..., vk are linearly independent then we can find a subset of v, v2,..., v n of the form v, v 2,..., v k, vi, vi2,..., v ir consisting of linearly independent elements whose linear span is also W. Carollary : If V is a finite dimensional vector space then it contains a finite set v, v 2,..., v n of linearly independent elements whose linear span is V. Basis : A subset S of a vector space V is called a basis of V if S consist of linearly independent elements and V = L (S). Carollary : If V is a finite dimensional vector space and if u, u 2,..., u m span V then some subset of u, u2,..., u m forms a basis of V.. Lemma : If v, v2,..., v n is a basis of V over F and if w, w2,..., w m in V are linearly independent over F, then m n. Carollary : If V is finite dimensional over F then any two basis of V have the same number of elements. Carollary : ( n) F is isomorphic to ( m) F if and only if n = m. (by above carollary). (Two vector spaces are isomorphic if and only if dimension is same). Carollary : If V is finite dimensional over F then V is isomorphic to ( n) F for a unique integer n. (integer n depends on dimensions of V). (No. of elements in the basis : dimension). 5

Carollary : Any two finite dimensional vector spaces over F of the Same dimension are isomorphic. Lemma : If V is finite dimensional over F and if u, u2,..., um V are linearly independnt then we can find vectors um+, um+ 2,..., um+ r V such that u, u 2,..., um, um+, u m+ rform a basis of V. Lemma : If V is finite dimensional and if W is a subspace of V then W is finite dimensional, V dimw dimv and dim dim( V) dim( W) W =. Carollary : If A and B are finite dimensional subspaces of a vector space V then A + B is finite dimesnional and dim( A+ B) = dima+ dimb dim ( A B ). Dual Space Lemma : Hom (V, W) is a vector space over F under the operation ( T+ S)( v) = T( v) + S( v ). T ( αv) = α T( v), α F and v V. Proof : Let V and W be vector spaces over F and consider a collection of homomorphisms form V to W. As, (Hom (V, W), +) (i) Let ( T+ S)( v) = T( v) + S( v) and T ( αv) = α T( v) ; where α F and v V. Take v, v 2 V and T, S Hom (V, W) ( T S)( v v ) T ( v v ) S( v v ) + + = + + + = T v + T v + S v + S v 6

= T v + S v + T v + S v 2 ( T S)( v ) ( T S)( v ) = + + + (, ) T+ S Hom VW (ii) Scalar Multiplication Let ( T+ S)( αv) = T ( αv) + S ( α v) ; α F and v V. = αt( v) + α S( v)... T, S Hom( VW, ) = α ( T( v) + S( v) ) = α ( T+ S)( v) (iii) Associative Property Let T, T 2, T 3 Hom (V, W) and v V. (( T T ) T )( v) ( T T )( v) T ( v) + + = + + 3 3 Now, = T ( v) + T ( v) + T ( v) 3 = T ( v) + ( T + T )( v) 3 ( T ( T T ))( v) = + + 3 ( T T )( v + v ) = ( T + ( T ))( v + v ) = T v + v + T v + v 2 = T v + T v + T v + T v 2 2 ( ) ( ) = T v + T v + T v + T v 2 2 = T v + T v T v T v 2 2 = T v T v + T v T v 2 2 ( T ( T ))( v ) ( T ( T ))( v ) = + + + 2 ( T T )( v ) ( T T )( v ) = + 2 7

, T T Hom VW Also ( T T )( αv) = T ( αv) + ( T )( αv) ( (, ), ) = αt ( v) + ( ) T ( αv) = αt ( v) αt ( v) = α T T v Hom VW + is a group. Let ( T + T )( v) = T ( v) + T ( v) (,,+) Hence, = T ( v) + T ( v ) T ( v ), T ( v) W and W is a vector S. 2 = ( T + T )( v) 2 Hom VW is abelian. Let λ F, T Hom( VW, ) and ab, F. ( T)( av bv ) T( ( av bv )) 2 λ + = λ +... by linearity of T. (, ) λt Hom VW ( λ ) ( λ ) = T av + T av = λat v + λbt ( v ) ( λ ) ( λ ) = a T v + b T v Scalar multiplication distrubute over addition. λ ( T T )( v) λ ( T T )( v) + = + = λ T v + T2 v ( T )( v) ( T )( v) = + v, v V. λ λ 2... W is a V. S. and (( λt ) ( λt ))( v) = + 8

Vector multiplication distrubute over scalar addition ( λ + β) T ( v) = ( λ + β) ( T ( v) ) = λt ( v) + β T ( v)... W is a Vector space = ( λt )( v) + ( βt )( v)... λt, βλ Hom( VW, ) = ( λt + βt )( v) T ( v) = ( T ( v) ) λβ λ β... T ( v) W = β ( λt( v) ) Identity w.r.t. multiplication T ( v) = T( v) = T ( v) = T v Hom (V, W) is a vector space over F. Hence the proof. Theorem : mn over F. Proof : If V and W are of dimensions m and n respectively over F, then Hom (V, W) is of dimension We prove the theorem by exhibiting a basis of Hom (V, W) over F consisting m, n elements. Let v, v 2,..., v m be a basis of V over F and w, w2,..., w n be a basis of W over F. Define Tij : V W as T ( v) = w ij i j We claim that (, ) and spans Hom (V, W). Let vu, V. ij α and = T v w ; for i = k. ij k j = 0 ; for i k. T Hom VW and { ij =,2,...,, =,2,..., } v= α v + α2v2 +... + α m v m and T i m j n is linearly independent u = β v + β2v2 +... + β m v m, α, α2,..., α m, β, β2,..., β m F. 9

Now, ( α + β ) = ( αα + αα +... + αα + ββ + ββ +... + ββ ) T v u T v v v v v v ij ij 2 m m 2 m m ( αα ββ ) ( αα ββ ) = T + v +... + + v ij m m m ( αα ββ ) = 0 +... + + w + 0 +... + 0 ( αα ββ ) = + i i j i i j w ( 0 w... αiwj 0 wj+... 0 wm) = α + + + + + ( 0 w... βiwj 0 wj... 0 wn) + β + + + + + = αt ( v) + βt ( u) ij (, ) T Hom VW. ij Let S (, ) S v W. Hence, Hom VW and v V. S v = αw + α2w2 +... + α n w n. For some α, α2,..., α n F. In fact, 2... Consider ij S v = α w + α w + + α w, i =, 2,..., m. i i i in n S = α T + α T +... + α T + α T +... + α T 0 2 2 n n 2n 2n Let us compute S0 ( v i ). +... + αmtm +... + α mn T mn [ α... α... α... α ] S v = T + + T + + mtm + + T v 0 i n n mn mn i = 0 +... + 0 + α w + α w +... + α w + 0 +... + 0 = S( vi ) i i2 2 in n Thus, the Homomorphisms S 0 and S agree on the basis of V. S0 = S 0 T ij ( v ) k wj i= k = 0 i k

However, S 0 is linear combination of T ij whence S must be the same linear combination. Thus, the set B spans Hom (V, W). Now, we will show B is linearly independent Suppose, ( T T T T T ) β + β +... + β +... + β +... + β = 0; 2 2 n n m m mn mn where B F. ij Apply this on a basis vector v i of V. ( β T β T β T β T β T )( v ) ( v ) + +... + +... + +... + = 0 = 0 2 2 n n m m mn mn i i 0 +... + 0 + β w + β w +... + β w + 0 +... + 0= 0 βi = βi2 =... β in. i i i2 2 in n Since w ' s are basis elements of W. i This implies β = 0 for all i and j. ij Thus, β is linearly independent over F and forms a basis of Hom (V, W) over F. dim Hom (V, W) is mn. Hence the proof. Corollary :. If dim V = m then dim Hom (V, V) = m 2. Proof : Replace W by V and n by m. 2. If dim V = m then dim (Hom (V, F)) = m. Proof : As F a vector space is of dimension one over F. Note : If V is finite dimensional over F. It is isomorphic to Hom (V, F). Dual Space : ˆ V If V is a vector space then it s dual space is Hom (V. F). It is denoted by V ˆ. The elements of will be called a linear functional on V into F.

Problem : Show that vˆ, vˆ 2,..., v ˆn is a basis of V ˆ, for v, v2,..., v n is basis of V and Solution : i( j) vˆ v = if i = j = 0 if i j.... () Consider α vˆ + α2vˆ2 +... + αnvˆn = 0 for α, α2,..., αn F ( v ˆ v ˆ )( v ) α +... + α = 0 α = 0 it is true i n n i i vˆ,..., vˆn are linearly independent. ˆ = ˆ ˆ ˆ is basis of V.. dimv dim V v, v2,..., vn Lemma : If V is finite dimensional and v 0 in V then there is an element f V ˆ such that f ( v ) 0. Proof : Let V is finite dimensional vector space over F and let v, v 2,..., vn be a basis of V.. Let vˆ i V ˆ defined by vˆ i( v j) = if i = j Consider, vˆ ( v ) = 0 for v 0 in V, i ( α α α ) vˆ v + v +... + v = 0 i n n n = 0 if i j. n v= αivi, i F i= α. α i =0... by definition of v ˆi. All Definition : α i ' s used in the representation of v are zero. Hence v = 0, a contradiction. Thus, vˆ ( v ) 0. i vˆ = f V ˆ such that f ( v ) 0. i Hence the proof. Let the functional on ˆ with T = ( f + g )( v ) v 0 0 V into F, Tv ( f ) f ( v0 ) 0 0 0 = for f V ˆ. = f v + g v... Hom (V, F) = T f + T g v v 0 0 2

( λ ) = λ T f f v v 0 0 v 0 λ T f T v 0 is in dual of V ˆ it is called Second Dual of V. It is denoted by V ˆ. Leema : If V is finite dimensional then there is an isomorphism of V onto ˆ V. Proof : Let V is finite dimensional vector space. Define the map ψ : V Vˆ by ψ ( v) = Tv for every v V. We will show ψ is well-defined, one-one, onto, homomorphism. Let uv, V. Let u = v. f ( u) = f ( v )... f Vˆ T = T u ψ ( u) = ψ ( v) v Thus ψ is well-defined and one-one. Now, consider uv, V. ( u + v) = Tu v ψ + but T u+ v = f u + v = + f u f v = T ( f ) + T ( f ) ( T T )( f ) = + u T = T + T u+ v u v ψ ( u + v) = Tu + Tv v = ψ ( u) + ψ ( v) ψ ( u + v) = ψ ( u) + ψ ( v) Let α be any element. ψ ( αv) = T = f ( αv) = α f ( v) αv u = α... Vector space. T v = αψ ( v) ψ is homomorphism. v 3

Annihilator : If W is a subspace of V then the annihilator of W { } AW = f V ˆ f ( w ) = 0; w W EXERCISE.... Show that AW is subspace of V ˆ. Proof : Let f, g A( w ) and α, β F ; w is arbitrary element in W. Claim : α f + βg AW f ( w) = 0 = g( w ), w W. ( α f + βg)( w) = α f ( w) + βg( w) = α 0+ β 0 = 0 α f + βg AW Hence the proof. Note :. If W = { 0} is the 0 subspace of V then AW = V ˆ. 2. If W = V then AW = { 0}. 3. If V is finite dimensional vector space and W contains a non-zero vector and also W is a proper subspace then AW is non-trivial, proper subspace of V ˆ. Lemma : If V is finite dimensional vector space over F and W is a subspace of V then W ˆ to V ˆ / AW and dim( AW ) = dimv dimw. Proof : Let W be a subspace of V where V is finite dimensional. is isomorphic If f V ˆ, let f be the restriction of f to W and is defined on W. As, f ( w) = f ( w ) for every w W. 4

f Wˆ ; since f V ˆ. Now, consider the mapping T : Vˆ W ˆ ; defines as = T f f ; for f V ˆ. Let f, g V ˆ such that f = g. f ( v) = g( v ); for every v V i.e. f ( v) = g( v ); for every v W V f ( v) = g( v ) ; for v W f = g T ( f ) = T( g )... by definition of T. Hence T is well-derined and one-one. For Homomosphism of T, Consider, T ( f + g) = f + g ( f + g)( v) = ( f + g)( v) = f ( v) + g( v) = ( f + g)( v) T ( f + g) = f + g = T( f ) + T ( g) and T ( f ) = f = f = T( f ) λ λ λ λ... since λ is scalar. T is Homomorphism. Now, we will show that T is onto. i.e. for a given any element h W ˆ. Then h is the restriction of some f V ˆ. i.e. f = h 5

dim (V) = n. We know that, if w, w2,..., w m is a basis of W, subspace of V.. Then it can be expanded to a basis of V of the form {,...,,... } Let W be the subspace of V spanned by {,,..., } Thus, = w w w. m+ m+ 2 n w w w w ; whose m m+ n V W W ( { w w } does not belong to { } and W W = whole space) m n +,..., Any element of V is represented as v= w + w ; w W and w W. For h W ˆ, define f V ˆ as f ( v) = h( w ). f w+ w = h w f w + f w = h w f ( v) = h( w) f Vˆ we have f = h. f ( w) = h( w) Thus, T ( f ) = h and so T maps V ˆ onto W ˆ. Consider, { ˆ } ker T = f V T f = 0 { f V ˆ f 0 } = = = { f V ˆ f ( w ) = 0 w W } w,..., w n so W W = φ = { f V ˆ f ( w ) = 0 w W }... f ( w ) = f ( w ) every w W. = AW Thus, by fundamental theorem of isomorphism (algebra). Wˆ Vˆ AW In particular they have the same dimensions ˆ dim ˆ dim V W = AW, 6

Also, we know Above expression become, dim dim V W = AW dimw = dimv dim AW Hence the proof. dimv = dimv ˆ and dimw = dimw ˆ. Theorem : Proof : If V be a finite dimensional vector space over the field F. Let W be a subspace of V, then dimw + dimaw = dimv If W is the 0 subspace of V then AW = V ˆ. dim( AW ) = dimvˆ = dim V. Similarly, the result is obvious when W = V. also. Let us suppose that W is proper subspace of V and dim W = m, dim V = n with 0 < m < n. Let {,,..., } B w w w be as basis for W. Since, B is linearly independent subset of V = m It can be extended to form a basis for V.. Let {,,...,,,..., } B w w w w w be a basis for V.. = m m+ n Let ˆ {,,...,,,..., } B f f f f f be a dual basis of V.. = m m+ n Then ˆB is a basis for ˆ V such that i j = ij = 0 We claim that {,,..., } Since = m+ m+ 2 n f w δ ; if i j = ; if i = j. S f f f is a basis of AW. S B ˆ. Therefore S is linearly independent because ˆB is linearly independent. Therefore, S is basis of AW if LS = AW. Let f AW, f V ˆ. 7

So, let f = n α ifi ; α F... () i= i Now, f AW. f ( w ) = 0 w AW. ( j ) 0 f w = for each j =, 2,...,m. ( w ) n αifi j i= = 0 ( j) ( j ) j j ( j) i i( j) n n( j) α f w + α f w +... + α f w + α f w +... + α f w = 0 2 α j =0 ( j) 0, f i w = i j =, i= j Putting, α = α2 =... = α = 0 in () m f = n αifi i= m+ f L( S) A( W ) contained in L( S ). Let g L( S ). = n g Bf i i Let i= m+ w W. w = m γ jwj j= n m g( w) = βifi γ jwj i= m+ j= m n = γ β f ( w ) j i i j j= i= m+ 8

( + f + ( w )... f ( w )) m γ j β m m j β n n j j= = + + +... γ β = γβ f w + + f w + m+ m+ m m m+ j = 0 γβ +... + γ β f w f w m+ 2 m+ 2 m m+ 2 m+ 2 m γ β +... + γβ f w +... + f w n n m n n n g( w) = 0 Hence, g AW. Therefore, L( S) AW we have LS = AW dimvˆ = dimaaw ( ) + dim AW dim ( AW ) = n m = dimv dimw dimv = dimw + dim AW Anihilator of Anihilator Let V be a vector space over F if any subset of V then A (S) is subspace of V ˆ definition of annihilator. { } A ( A ( S )) = L V ˆ L ( f ) = 0, f A ( S ) and by Example : ) Show that A( A( S )) is subspace of V ˆ. Note : v L v. If V is finite dimensional vector space then we have identity ˆ V with V through the isomorphism 9

Therefore, we may regard A( A( S )) as subsapce of V.. { } A( A( S) ) = v V f ( v) = 0 f AS Corollary : If W is subspace of V finite dimensional vector space then AAW ( ) = W. Proof : We have, { } AW = f V ˆ f ( w ) = 0, w W... () { } AAW ( ) = v V f ( v) = 0, f AW... (2) Let w W. Then by () f ( w ) = 0, w W V. There from equation (2); f ( w ) = 0, f AW. Therefore we have, AAW ( ) Hence, W AAW Let v AAW, v V. f ( v ) = 0, f AW f ( v ) = 0, f AW for v W AAW ( ) W. AAW ( ) = W dimvˆ = dimaaw ( ) + dim AW dimv = dimaaw ( ) + dim( V) dim W dimaaw ( ) + dim( W) AAW ( ) = W. Hence the required. Problem : Let V be finite dimensional vector space over the field F. If S is subset of V prove that A( S) = A L( S ) where L( S ) is linear span of S. Solution : Let V be a finite dimensional and S is any subset of V. We know that S L( S ). 20

Therefore, A L( S) AS... () ( W V and W = { 0} V AW = Vˆ, AV = { 0}, V ˆ { 0 } ) Now, let f A( S ) then f ( s ) = 0 for all s S. If u is any element of L( S ) then u = α isi L S. Consider, = n f u f αisi i= i= = n αi f Si n i= n i= = α i 0; S S i = 0 f A( L( S ))... (2) A( S) A L( S) A( L( S) ) = A( S) Hence the result. Problem : Let V be finite dimensional vector space over F. If S is any subset of V then prove that A( A( S) ) = L( S ). Solution : By previous A( S) = A L( S ). By taking annihilator on both sides, A( A( S) ) = L( S )... L( S ) is subspace of V.. Problem : Let W and 2 terms of AW ( ) and ( 2 ) Solution : W be subspaces of V which is finite dimensional. Describe ( + ) AW. Let W and W 2 be two subspaces of V. AW W in 2

We have, W W + W 2 and W2 W + W 2. Since, AW ( + W ) AW and ( + ) AW + W AW AW Conversely let, f AW AW f AW and f AW f ( w ) = 0, Let any f w 2 = 0. 2 AW W AW. 2 w W, w2 W 2. w W. Thus w W is represented as w= w+ w 2. f w = f w + w = f w + f w = 0 f ( w) = 0 f AW + W AW AW AW + W AW + W = AW AW Hence, the result. Problem : If W and W 2 be subspaces of finite dimensional vector spaces. Describe AW ( W ) in terms of + AW AW. Solution : By using previous exercise by replacing V by ˆ V, ( + ) = ( ) ( ) AAW AW AAW AAW = W W ( ( ( ) ( 2) )) ( ) A A A W + AW = AW W W by AW, 2 AW AW AW ( W )... + + = Hence, the result. W by AW we get; AW AW 2 is a subspace of V ˆ. 2 22

System of Linear Homogenous Equation Theorem : If the system of homogeneous linear equations a x + a x +... + a x = 0 n 2 2 n n a x + a x +... + a x = 0 2 2 2 2n n amx+ am2x2 +... + amnxn = 0 where a ij F is of rank r then there are n r linearly independent solutions in ( n) F. Proof : Consier, the system of m equations and n unknowns. a x + a x +... + a x = 0 2 2 n n amx+ am2x2 +... + amnxn = 0 where a ij F Now, we find how many linearly independent solutions ( x, x 2,..., x n) in ( n) F. Let U be the subspace generated by m vectors with ( a a a ),,..., n 2 ( a a a ),,..., n 22 2 (,,..., ) a a a and supposed that U is of dimensions r. m m2 mn Let v =, v =,..., = ( 0,0,0,...,0,),0,...,0 2 0,,...,0 v n be a basis for ( n) F. vˆ, vˆ2,..., v ˆn be it s dual basis of ˆ ( n ) F, any small f F ˆ n can be expressed as a linear combination of vˆ i ' s. = n f xv iˆi ; x F i= a, a,..., a U. For 2 n i We have, f ( a, a,..., a ) = f ( a v + a v +... + a v ) 2 n 2 2 n n ( a (,0,...,0) + a ( 0,,...,0 ) = ( a, a,...)) 2 2 23

... = f a v + f a v + + f a nv 2 2 n n n... = a xvˆ v + a xvˆ v + + a xvˆ v i i 2 i i 2 n i i n i= i= i= = ax + a2x2 +... + a n x n ( j)... v ˆ v = 0 i i j = 0 This is true for the other vectors in U. f AU = i= j Every solution ( x, x 2,..., x n) of the system of homogenous equation it s an elements xv ˆ + xv 2ˆ2+... + xv n ˆ n in AU. Therefore, we see that the number of linearly independent solutions of the system of equation is the dimension of AU. But we know, ( n) dimf = dimu + dim AU dim AU = n r Hence, the proof. Corollary : If n > m that is if no. of unknowns exceeds the number of equations then there is a solution ( x, x 2,..., x n) where not all of ( x, x 2,..., x n) are 0. Proof : Since U is generated by m vectors and m < n also r = dim( U) < m. By above theorem the dim AU = n r this number is nothing but no. of elements in the basis of AU which are non-zero vectors. Hence, the proof. 24

EXERCISE :. If ST, Hom( VW, ) and S( v ) T( v ) S = T. i = for all elements v i of a basis of V, prove that i 2. If V is finite dimensional and v v 2 are in V, prove that there is an f Vˆ such that f ( v ) f v. 3. If F is the field of real numbers, find A (W), where W is spanned by (, 2, 3) and (0, 4, ). 4. If f and g are in V ˆ such that f ( v ) = 0 implies g( v ) = 0, prove that g = λ f for some λ F. 25