Abstract¶ The conditions for a set of vectors to be a basis and some properties of vector basis.
Keywords: Vectors Vector Spaces basis ¶ Basis Conditions ¶ Let V V V a 1 , a 2 , … , a n ∈ V ∀ n a_{1}, ~a_{2}, \ldots, ~a_{n}~\in~V~\forall~n a 1 , a 2 , … , a n ∈ V ∀ n
The set of vectors { a 1 , a 2 , … , a n } \{ a_{1}, ~a_{2}, \ldots, ~a_{n} \} { a 1 , a 2 , … , a n } V V V
Span ( { a 1 , a 2 , … a n } ) = V \textrm{Span}(\{ a_{1}, ~a_{2}, \ldots ~a_{n} \}) = V Span ({ a 1 , a 2 , … a n }) = V
Any vectors in V V V { a 1 , a 2 , … a n } \{ a_{1}, ~a_{2}, \ldots ~a_{n} \} { a 1 , a 2 , … a n } 💭
The vectors { a 1 , a 2 , … a n } \{ a_{1}, ~a_{2}, \ldots ~a_{n} \} { a 1 , a 2 , … a n } linear independent
Use of Basis ¶ For any ψ ∈ V \psi~\in~V ψ ∈ V { a 1 , a 2 , … , a n } \{ a_{1}, ~a_{2}, \ldots, ~a_{n} \} { a 1 , a 2 , … , a n } λ 1 , λ 2 , … , λ n ∈ F n \lambda_{1}, \lambda_2, \ldots, \lambda_{n}~\in~\mathbb{F}^{n} λ 1 , λ 2 , … , λ n ∈ F n
ψ = λ 1 a 1 + λ 2 a 2 + … + λ n a n . \psi = \lambda_{1} a_{1} + \lambda_2 a_{2} + \ldots + \lambda_{n} a_{n}. ψ = λ 1 a 1 + λ 2 a 2 + … + λ n a n . Hence, given a basis one can write any vector in the space as a sum over those vectors in the basis.
Dimension Definition ¶ The dimension of a vector space V V V
Therefore, if a vector space is known to be of dimension n n n n n n
Basis Properties ¶ Special Basis ¶
let V V V F n \mathbb{F}^{n} F n
The standard basis of a vector space V V V
e 1 = [ 1 0 ⋮ 0 ] , e 2 = [ 0 1 ⋮ 0 ] , e n = [ 0 0 ⋮ 1 ] . \begin{align*}
e_{1} &= \begin{bmatrix}
1 \\
0 \\
\vdots \\
0
\end{bmatrix}, ~ ~
e_2 &= \begin{bmatrix}
0 \\
1 \\
\vdots \\
0
\end{bmatrix}, ~~
e_n &= \begin{bmatrix}
0 \\
0 \\
\vdots \\
1
\end{bmatrix}.
\end{align*} e 1 = ⎣ ⎡ 1 0 ⋮ 0 ⎦ ⎤ , e 2 = ⎣ ⎡ 0 1 ⋮ 0 ⎦ ⎤ , e n = ⎣ ⎡ 0 0 ⋮ 1 ⎦ ⎤ . Let the set of vectors { a 1 , a 2 , … , a n } \{ a_{1}, ~a_{2}, \ldots, ~a_{n} \} { a 1 , a 2 , … , a n } V V V
a i ⋅ a j = δ i j , a_{i} \cdot a_{j} = \delta_{ij}, a i ⋅ a j = δ ij , where δ i j \delta_{ij} δ ij Kronecker delta function given by
δ i j ≔ { 1 if i = j 0 if i ≠ j \delta_{ij} \coloneqq \left\{
\begin{array}{ll}
1 & \mbox{if } i = j \\
0 & \mbox{if } i \neq j
\end{array}
\right. δ ij : = { 1 0 if i = j if i = j Tensor Product Basis ¶ Let the set of vectors { a i } i \{ a_i \}_i { a i } i A A A { b j } j \{ b_j \}_j { b j } j B B B
The set of vectors { a i ⊗ b j } i , j \{ a_i \otimes b_j\}_{i,j} { a i ⊗ b j } i , j A ⊗ B A \otimes B A ⊗ B ⊗ \otimes ⊗ tensor product .
A general element of the space A ⊗ B A \otimes B A ⊗ B U U U
U = ∑ k c k v k ⊗ u k , U = \sum_k c_k v^k \otimes u^k, U = k ∑ c k v k ⊗ u k , where
c k ∈ C , v k ∈ A , u k ∈ B ∀ k . c_k \in \mathbb{C}, ~ ~ v^k \in A, ~ ~ u^k \in B ~ ~\forall~k. c k ∈ C , v k ∈ A , u k ∈ B ∀ k . Given v k ∈ A v_k \in A v k ∈ A { a i } i \{ a_i \}_i { a i } i A A A { α i k : α i k ∈ C } i \{\alpha^k_i : \alpha^k_i \in \mathbb{C} \}_i { α i k : α i k ∈ C } i
v k = ∑ i α i k a i . v^k = \sum_i \alpha^k_i a_i. v k = i ∑ α i k a i . The same holds for u k ∈ B u^k \in B u k ∈ B
u k = ∑ i β i k b i , u^k = \sum_i \beta^k_i b_i, u k = i ∑ β i k b i , where { β i k : β i k ∈ C } i \{\beta^k_i : \beta^k_i \in \mathbb{C} \}_i { β i k : β i k ∈ C } i
Hence, U U U
U = ∑ k ∑ i ∑ j c k α i k β j k a i ⊗ b j = ∑ i ∑ j κ i , j a i ⊗ b j , \begin{align*}
U &= \sum_k \sum_i \sum_j c_k \alpha^k_i \beta^k_j a_i \otimes b_j \\
&= \sum_i \sum_j \kappa_{i,j} a_i \otimes b_j,
\end{align*} U = k ∑ i ∑ j ∑ c k α i k β j k a i ⊗ b j = i ∑ j ∑ κ i , j a i ⊗ b j , where κ i , j = ∑ k c k α i k β j k \kappa_{i,j} = \sum_k c_k \alpha^k_i \beta^k_j κ i , j = ∑ k c k α i k β j k A ⊗ B A \otimes B A ⊗ B { a i ⊗ b j } i , j \{ a_i \otimes b_j\}_{i,j} { a i ⊗ b j } i , j
This shows that { a i ⊗ b j } i , j \{ a_i \otimes b_j\}_{i,j} { a i ⊗ b j } i , j A ⊗ B A \otimes B A ⊗ B
To be a basis, the vectors { a i ⊗ b j } i , j \{ a_i \otimes b_j\}_{i,j} { a i ⊗ b j } i , j
dim ( A ⊗ B ) = dim ( A ) × dim ( B ) . \textrm{dim}(A \otimes B) = \textrm{dim}(A) \times \textrm{dim}(B). dim ( A ⊗ B ) = dim ( A ) × dim ( B ) . Then, as
dim ( A ) = ∣ { a i } i ∣ , dim ( B ) = ∣ { b i } i ∣ , \textrm{dim}(A) = \vert \{ a_i \}_i \vert, ~ ~ \textrm{dim}(B) = \vert \{ b_i \}_i \vert, dim ( A ) = ∣ { a i } i ∣ , dim ( B ) = ∣ { b i } i ∣ , it is the case that
∣ { a i ⊗ b j } i , j ∣ = dim ( A ⊗ B ) . \vert \{ a_i \otimes b_j \}_{i,j} \vert = \textrm{dim}(A \otimes B). ∣ { a i ⊗ b j } i , j ∣ = dim ( A ⊗ B ) . Hence, the set of vectors { a i ⊗ b j } i , j \{ a_i \otimes b_j \}_{i,j} { a i ⊗ b j } i , j A ⊗ B A \otimes B A ⊗ B A ⊗ B A \otimes B A ⊗ B