Vector Analysis HOMEWORK IX Solution. 1. If T Λ k (V ), v 1,..., v k is a set of k linearly dependent vectors on V, prove

Transcription

1 1. If T Λ k (V ), v 1,..., v k is a set of k linearly dependent vectors on V, prove T ( v 1,..., v k ) = 0 Since v 1,..., v k is a set of k linearly dependent vectors, there exists a 1,..., a k F such that a 1 v a k v k = 0 and at least one 0. Then it implies we can write So T ( v 1,..., v n ) =T ( v 1,..., v i,..., v k ) v i = a 1 v =T ( v 1,..., a 1 v v i 1 +1 v i 1 +1 v i+1... a k v k v i+1... a k v k,..., v k ) = a 1 T ( v 1,... v i 1, v 1, v i+1,..., v k )... 1 T ( v 1,... v i 1, v i 1, v i+1,..., v k ) +1 T ( v 1,... v i 1, v i+1, v i+1,..., v k )... a k T ( v 1,... v i 1, v k, v i+1,..., v k ) = =0. If T Λ k (V ) and S Λ l (V ), prove T S = ( 1) kl S T If { v 1,..., v n } is a basis for V and φ 1,..., φ n is the dual basis for V. First we show that φ i φ j = φ j φ i : φ 1 φ j ( u 1, u ) = σ S sng(σ)φ i ( u σ(1) )φ j ( u σ() ) =φ i ( u 1 )φ j ( u ) φ i ( u )φ j ( u 1 ) = φ j ( u 1 )φ i ( u ) + φ j ( u )φ i ( u 1 ) = σ S sgn(σ)φ j ( u σ(1) )φ i ( u σ() ) = φ j φ i ( u 1, u ) 1

2 This implies φ i1... φ ik φ j1... φ jl = ( 1) kl φ j1... φ jl φ i1... φ ik If T = 1 i 1 <...<i k n T i 1...i k φ i1... φ ik and S = 1 j 1 <...<j l n S j 1...j l φ j1... φ jl, we get T S =( = 1 i 1 <...<i k n 1 i 1 <...<i k n 1 j 1 <...<j l n =( 1) kl =( 1) kl =( 1) kl ( =( 1) kl S T φ i1... φ ik ) ( 1 i 1 <...<i k n 1 j 1 <...<j l n 1 j 1 <...<j l n φ j1... φ jl ) φ i1... φ ik φ j1... φ jl 1 j 1 <...<j l n 1 i 1 <...<i k n 1 j 1 <...<j l n φ j1... φ jl φ i1... φ ik φ j1... φ jl φ i1... φ ik φ j1... φ jl ) ( 1 i 1 <...<i k n φ i1... φ ik ) 3. If { v 1,..., v n } is a basis for a n-dimensional vector space V and {φ 1,..., φ n } is the dual basis for V, prove for each σ S n, φ σ(1)... φ σ(n) = sgn(σ)φ 1... φ n Method I: By (), φ i φ j = φ j φ i. We can get from φ σ(1)... φ σ(n) to φ 1... φ n by odd number of transpositions if sgn(σ) = 1, and even number of transpositions if sgn(σ) = +1, so we get the conclusion.

3 Method II: φ σ(1)... φ σ(n) ( v 1,..., v n ) =n!alt(φ σ(1)... φ σ(n) )( v 1,..., v n ) = sgn(τ)φ σ(1) ( v τ(1) )...φ σ(n) ( v τ(n) ) τ S n = sgn(τσ)φ σ(1) ( v τσ(1) )...φ σ(n) ( v τσ(n) ) τ S n =sgn(σ) τ S n sgn(τ)φ σ(1) ( v τ(σ(1)) )...φ σ(n) ( v τ(σ(n)) ) =sgn(σ) τ S n sgn(τ)φ 1 ( v τ(1) )...φ n ( v τ(n) ) =sgn(σ)n!alt(φ 1.. φ n )( v 1,..., v n ) =sgn(σ)φ 1... φ n ( v 1,..., v n ) 4. T T k (V ) is called a symmetric k-tensor if for any v 1,..., v k V and any σ S k, T ( v σ(1),..., v σ(k) ) = T ( v 1,..., v k ). The set of all symmetric k-tensors form the subspace Sym k (V ) of T k (V ). (i). Given any T T k (V ), define Sym(T ) to be the k-tensor: Prove Sym(T ) Sym k (V ). For any τ S k, Sym(T )( v 1,..., v k ) = 1 T ( v σ(1),..., v σ(k) ) σ S k Sym(T )( v τ(1),..., v τ(k) ) = 1 T ( v στ(1),..., v στ(1) ) σ S n = 1 T ( v σ(1),..., v σ(1) ) σ S n =Sym(T )( v 1,..., v k ) 3

4 The second last equality is because for fixed τ S k, there is a bijection given by σ στ S k S k (ii). If T Sym k (V ), prove Sym(T ) = T If T Sym k (V ), Sym(T )( v 1,..., v k ) = 1 T ( v σ(1),..., v σ(k) ) σ S k = 1 σ S k T ( v 1,..., v k ) = 1 T ( v 1,..., v k ) =T ( v 1,..., v k ) (iii). T T (V ), prove T = Sym(T ) + Alt(T ) If T T (V ), So Sym(T )( v 1, v ) = 1! Alt(T )( v 1, v ) = 1! T ( v σ(1), v σ() ) = T ( v 1, v ) + T ( v, v 1 ) σ S sgn(σ)t ( v σ(1), v σ() ) = T ( v 1, v ) T ( v, v 1 ) σ S [Sym(T ) + Alt(T )]( v 1, v ) =Sym(T )( v 1, v ) + Alt(T )( v 1, v ) = T ( v 1, v ) + T ( v, v 1 ) =T ( v 1, v ) + T ( v 1, v ) T ( v, v 1 ) 4

5 We conclude T = Sym(T ) + Alt(T ) (iv). Prove Sym (V ) Λ (V ) = {0} If T Sym (V ), then T ( v 1, v ) = T ( v, v 1 ). If T Λ (V ), then T ( v 1, v ) = T ( v, v 1 ). Combining these two equations, we see if T Sym (V ) Λ (V ), then T ( v 1, v ) = 0 for any v 1, v V. (v). T T (V ). Prove T can be decomposed as the sum of a symmetric - tensor and an alternating -tensor in a unique way. (Remark: This is equivalent to say T (V ) = Sym (V ) Λ (V )) By (iii), we know T = Sym(T ) + Alt(T ), which gives one such decomposition. We next prove this decomposition is unique: If T = ω 1 + η 1 = ω + η, where ω 1, ω Sym (V ) and η 1, η Λ (V ), then ω 1 ω = η η 1 The left side belongs to Sym (V ) while the right side belongs to Λ (V ), so ω 1 ω = η η 1 Sym (V ) Λ (V ) = {0} We conclude ω 1 = ω and η 1 = η, so the decomposition is unique. (vi). { u 1,..., u n } is a basis of V. T T (V ). A = (j ) is the n n matrix such that j = T ( u i, u j ). A square matrix A is called skew-symmetric if A t = A, where A t denotes the transpose of A. Prove T Sym (V ) if and only if A is a symmetric matrix, and T Λ (V ) if and only if A is a skew-symmetric matrix. T Sym (V ) T ( e i, e j ) = T ( e j, e i ) j = a ji A is symmetric. T Λ (V ) T ( e i, e j ) = T ( e j, e i ) j = a ji A is skewsymmetric. (vii). Prove Each n n matrix can be written as a sum of a symmetric matrix and a skew-symmetric matrix in a unique way. Fixing a basis { e 1,..., e n }, there is a one-to-one correspondence between -tensors on V and n n matrices as described in (vi). So given n n matrix A, it corresponds to some -tensor T. In (v) we proved T = ω + η can be decomposed as a sum of a symmetric -tensor ω and an alternating -tensor 5

6 η in a unique way, and in (vi) we proved ω Sym (V ) if and only if its matrix B is a symmetric matrix, and η Λ (V ) if and only if its matrix C is a skew-symmetric matrix, so A = B + C can be written as a sum of a symmetric matrix and a skew-symmetric matrix in a unique way. 1 3 (viii). Write the matrix as the sum of a symmetric matrix and a skew-symmetric matrix. If the matrix of T is A = (T ( e i, e) j ), then the matrix for Sym(T ) is ( T ( e i, e) j +T ( e j, e) i ) = 1(A + At ) and the matrix for Alt(T ) is ( T ( e i, e) j T ( e j, e) i ) = 1 (A At ). So A = 1(A + At ) + 1(A At ). Applying this formula we get =