Answer to Question #127665 in Linear Algebra for Bibhuti bhusan dehury

Question #127665
Show that C^n is a complex vector space
1
Expert's answer
2020-07-28T18:52:10-0400

We have to show that "\\mathbb{C}^n" is a complex vector space.


According to Wikipedia's definition of the vector space we can conclude that:

A vector space over a field "\\mathbb{F}" - is a set "V" together with two operations:

  • vector addition ( "+" ) : "V + V \\mapsto V" - takes any two vectors "v" and "w" and assigns to them a third vector which is commonly written as "v + w", and called the sum of these two vectors. ( The resultant vector is also an element of the set "V" )
  • scalar multiplication ( "\\cdot" ) : "\\mathbb{F} \\cdot V \\mapsto V" - takes any scalar "\\alpha" and any vector "\\upsilon" and gives another vector "\\alpha \\cdot v". ( Similarly, the vector "\\alpha \\cdot v" is an element of the set "V" ) ( Scalar multiplication is a multiplication of a vector by a scalar )

that satisfy the eight axioms listed below ( considering that "u, v, w" are arbitrary vectors in "V", and "\\alpha, \\beta" scalars in "\\mathbb{F}" ):

  1. associativity of addition : "u + (v + w) = (u + v) + w"
  2. commutativity of addition : "u + v = v + u"
  3. identity element of addition ( there exists an element "0 \\in V", called the zero vector, such that "v + 0 = v" for all "v \\in V" )
  4. inverse element of addition ( for every "v \\in V", there exists an element "-v \\in V" , called the additive inverse of "v" , such that "v + (-v) = 0" )
  5. associativity of multiplication : "\\alpha \\cdot (\\beta \\cdot v) = (\\alpha \\cdot \\beta) \\cdot v"
  6. identity element of scalar multiplication : "1 \\cdot v = v" ( where "1" denotes the multiplicative identity in "\\mathbb{F}" )
  7. distributivity ( of scalar multiplication with respect to vector addition ) : "\\alpha \\cdot (u + v) = \\alpha \\cdot u + \\alpha \\cdot v"
  8. distributivity ( of scalar multiplication with respect to field addition ) : "(\\alpha + \\beta) \\cdot v = \\alpha \\cdot v + \\beta \\cdot v"


Elements of "V" are commonly called vectors. Elements of "\\mathbb{F}" are commonly called scalars.


So, if we prove this 2 operation properties and 8 axioms for our "\\mathbb{C}^n" we will prove that it is a complex vector space.


VECTOR ADDITION

Let's consider "\\mathbb{C}^1" case:

Taking 2 elements from "\\mathbb{C}^1"-space: "u = a+ib \\in \\mathbb{C}^1" , and "v = c+id \\in \\mathbb{C}^1" , we can write: "u + v = a+ib + c+id = (a+b) + i(c+d) = t \\in \\mathbb{C}^1."

Now consider "\\mathbb{C}^n" case:

Taking 2 elements from "\\mathbb{C}^n"-space: "u = (a_1+ib_1, a_2+ib_2, \\ldots, a_n+ib_n) \\in \\mathbb{C}^n" , and "v = (c_1+id_1, c_2+id_2, \\ldots, c_n+id_n) \\in \\mathbb{C}^n" , we can write: "u + v = (a_1+ib_1, a_2+ib_2, \\ldots, a_n+ib_n) + (c_1+id_1, c_2+id_2, \\ldots, c_n+id_n) = \\newline \n((a_1+b_1) + i(c_1+d_1), (a_2+b_2) + i(c_2+d_2), \\ldots, (a_n+b_n) + i(c_n+d_n)) = t \\in \\mathbb{C}^n."

"\\square"


SCALAR MULTIPLICATION

The same way, let's consider "\\mathbb{C}^1" case:

Taking element from "\\mathbb{C}^1"-space: "u = a+ib \\in \\mathbb{C}^1" , and a scalar "\\alpha \\in \\mathbb{R}" , we can write: "\\alpha \\cdot u = \\alpha \\cdot (a+ib) = \\alpha \\cdot a + i \\alpha \\cdot b = t \\in \\mathbb{C}^1."

Now consider "\\mathbb{C}^n" case:

Taking element from "\\mathbb{C}^n"-space: "u = (a_1+ib_1, a_2+ib_2, \\ldots, a_n+ib_n) \\in \\mathbb{C}^n" , and a scalar "\\alpha \\in \\mathbb{R}" , we can write:

"\\alpha \\cdot u = \\alpha \\cdot (a_1+ib_1, a_2+ib_2, \\ldots, a_n+ib_n) = \\newline \n(\\alpha \\cdot (a_1+ib_1), \\alpha \\cdot (a_2+ib_2), \\ldots, \\alpha \\cdot (a_n+ib_n) ) = \\newline \n(\\alpha \\cdot a_1 + i \\alpha \\cdot b_1, \\alpha \\cdot a_2 + i \\alpha \\cdot b_2, \\ldots, \\alpha \\cdot a_n + i \\alpha \\cdot b_n)= t \\in \\mathbb{C}^n."

"\\square"


1) ASSOCIATIVITY OF ADDITION

Let's take 3 elements "u, v, w" from "\\mathbb{C}^n" -space: "u = (a_1+ib_1, \\ldots, a_n+ib_n)", "v = (c_1+id_1, \\ldots, c_n+id_n)", and "w = (f_1+ig_1, \\ldots, f_n+ig_n)", and show that "u + (v + w) = (u + v) + w."

"u + (v + w) = \\newline \n(a_1+ib_1, \\ldots, a_n+ib_n) + \\Big( (c_1+id_1, \\ldots, c_n+id_n) + (f_1+ig_1, \\ldots, f_n+ig_n) \\Big) = (a_1+ib_1, \\ldots, a_n+ib_n) + \\Big( (c_1+f_1)+i(d_1+g_1), \\ldots, (c_n+f_n)+i(d_n+g_n)\\Big) = \\newline \n\\Big( (a_1+c_1+f_1)+i(b_1+d_1+g_1), \\ldots, (a_n+c_n+f_n)+i(b_n+d_n+g_n)\\Big) = \\newline \n\\Big( (a_1+c_1)+i(b_1+d_1), \\ldots, (a_n+c_n)+i(b_n+d_n)\\Big) + (f_1+ig_1, \\ldots, f_n+ig_n) = \\newline \n\\Big( (a_1+ib_1, \\ldots, a_n+ib_n) + (c_1+id_1, \\ldots, c_n+id_n) \\Big) + (f_1+ig_1, \\ldots, f_n+ig_n) = \\newline \n(u + v) + w."

"\\square"


2) COMMUTATIVITY OF ADDITION

Let's take 2 elements "u, v" from "\\mathbb{C}^n"-space: "u = (a_1+ib_1, \\ldots, a_n+ib_n)", and "v = (c_1+id_1, \\ldots, c_n+id_n)", and show that "u + v = v + u."

"u + v = (a_1+ib_1, \\ldots, a_n+ib_n) + (c_1+id_1, \\ldots, c_n+id_n) = \\newline \n\\Big( (a_1+c_1)+i(b_1+d_1), \\ldots, (a_n+c_n)+i(b_n+d_n)\\Big) = \\newline \n\\Big( (c_1+a_1)+i(d_1+b_1), \\ldots, (c_n+a_n)+i(d_n+b_n)\\Big) = \\newline \n(c_1+id_1, \\ldots, c_n+id_n) + (a_1+ib_1, \\ldots, a_n+ib_n) = v + u."

"\\square"


3) IDENTITY ELEMENT OF ADDITION

There exist "0 = (0_1, 0_2, \\ldots, 0_n) \\in \\mathbb{C}^n", such that using VECTOR ADDITION property (proven above) we can write "v + 0 = v, \\forall v \\in V."

"\\square"


4) INVERSE ELEMENT OF ADDITION

For every "v = (c_1+id_1, \\ldots, c_n+id_n) \\in V", there exist element "-v = (-c_1-id_1, \\ldots, -c_n-id_n)", such that "v + (-v) = 0" (according to VECTOR ADDITION property).

"\\square"


5) ASSOCIATIVITY OF MULTIPLICATION

For any "v = (c_1+id_1, \\ldots, c_n+id_n) \\in V", and 2 scalars "\\alpha" and "\\beta", let's prove that: "\\alpha \\cdot (\\beta \\cdot v) = (\\alpha \\cdot \\beta) \\cdot v"

"\\alpha \\cdot (\\beta \\cdot v) = \\alpha \\cdot \\Big(\\beta \\cdot (c_1+id_1, \\ldots, c_n+id_n)\\Big) = \\newline \n\\alpha \\cdot \\Big(\\beta \\cdot c_1+i\\beta \\cdot d_1, \\ldots, \\beta \\cdot c_n+i\\beta \\cdot d_n\\Big) = \\newline \n\\Big(\\alpha \\cdot \\beta \\cdot c_1+i\\alpha \\cdot \\beta \\cdot d_1, \\ldots, \\alpha \\cdot \\beta \\cdot c_n+i\\alpha \\cdot \\beta \\cdot d_n\\Big) = \\newline \n\\Big( (\\alpha \\cdot \\beta) \\cdot c_1+(\\alpha \\cdot \\beta) \\cdot id_1, \\ldots, (\\alpha \\cdot \\beta) \\cdot c_n+(\\alpha \\cdot \\beta) \\cdot id_n \\Big) = \\newline.\n(\\alpha \\cdot \\beta) \\cdot (c_1+id_1, \\ldots, c_n+id_n) = (\\alpha \\cdot \\beta) \\cdot v"

"\\square"


6) IDENTITY ELEMENT OF SCALAR MULTIPLICATION

There exist "1 = (1_1, 1_2, \\ldots, 1_n) \\in \\mathbb{C}^n", such that using VECTOR ADDITION property (proven above) we can write "1 \\cdot v = v, \\forall v \\in V."

"\\square"


7) DISTRIBUTIVITY ( OF SCALAR MULTIPLICATION WITH RESPECT TO VECTOR ADDITION )

For any 2 elements "u, v" from "\\mathbb{C}^n"-space: "u = (a_1+ib_1, \\ldots, a_n+ib_n)", "v = (c_1+id_1, \\ldots, c_n+id_n)" and a scalar "\\alpha" let's prove, that "\\alpha \\cdot (u + v) = \\alpha \\cdot u + \\alpha \\cdot v".

"\\alpha \\cdot (u + v) = \\alpha \\cdot \\Big( (a_1+ib_1, \\ldots, a_n+ib_n) + (c_1+id_1, \\ldots, c_n+id_n) \\Big) = \\newline\n\\alpha \\cdot \\Big( (a_1+c_1)+i(b_1+d_1), \\ldots, (a_n+c_n)+i(b_n+d_n)\\Big) = \\newline \n\\Big( \\alpha \\cdot (a_1+c_1)+i\\alpha \\cdot (b_1+d_1), \\ldots, \\alpha \\cdot (a_n+c_n)+i\\alpha \\cdot (b_n+d_n)\\Big) = \\newline \n\\Big( (\\alpha \\cdot a_1+\\alpha \\cdot c_1)+i(\\alpha \\cdot b_1+\\alpha \\cdot d_1), \\ldots, \\alpha \\cdot a_n+\\alpha \\cdot c_n)+i(\\alpha \\cdot b_n+\\alpha \\cdot d_n) \\Big) = \\newline \n\\Big( (\\alpha \\cdot a_1)+i(\\alpha \\cdot b_1), \\ldots, \\alpha \\cdot a_n+i(\\alpha \\cdot b_n) \\Big) + \\newline\n \\Big( (\\alpha \\cdot c_1)+i(\\alpha \\cdot d_1), \\ldots, \\alpha \\cdot b_n+i(\\alpha \\cdot d_n) \\Big) = \\newline\n\\alpha \\cdot (a_1+ib_1, \\ldots, a_n+ib_n) + \\alpha \\cdot (c_1+id_1, \\ldots, c_n+id_n) = \\alpha \\cdot u + \\alpha \\cdot v."

"\\square"


8) DISTRIBUTIVITY ( OF SCALAR MULTIPLICATION WITH RESPECT TO FIELD ADDITION )

For any element "v = (c_1+id_1, \\ldots, c_n+id_n) \\in V", and 2 scalars "\\alpha" and "\\beta" let's prove that: "(\\alpha + \\beta) \\cdot v = \\alpha \\cdot v + \\beta \\cdot v."

"(\\alpha + \\beta) \\cdot v = (\\alpha + \\beta) \\cdot (c_1+id_1, \\ldots, c_n+id_n) = \\newline\n\\Big( (\\alpha + \\beta) \\cdot (c_1+id_1), \\ldots, (\\alpha + \\beta) \\cdot (c_n+id_n) \\Big) = \\newline\n\\Big( \\alpha \\cdot (c_1+id_1) + \\beta \\cdot (c_1+id_1), \\ldots, \\alpha \\cdot (c_n+id_n) + \\beta \\cdot (c_n+id_n) \\Big) = \\newline\n\\Big( \\alpha \\cdot (c_1+id_1), \\ldots, \\alpha \\cdot (c_n+id_n) \\Big) + \\Big( \\beta \\cdot (c_1+id_1), \\ldots, \\beta \\cdot (c_n+id_n) \\Big) = \\newline\n\\alpha \\cdot (c_1+id_1, \\ldots, c_n+id_n) + \\beta \\cdot (c_1+id_1, \\ldots, c_n+id_n) = \\alpha \\cdot v + \\beta \\cdot v."

"\\square"


"\\blacksquare"


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