Verify that each of the sets in Examples 1– 4 satisfies the axioms fora vector space. Find a basis for each of the vector spaces inExamples 1–4. I EXAMPLE 1 The set R" = {(a,, a, ..., a,) | a; E R} is a vector spaceover R. Here the operations are the obvious ones:%3D(a,, a, ..., a) + (b,, b» . .. , b,) = (a, + b,, a, + b» . .., a, + b,)1'andb(a,, az, . .. , a) = (ba,, ba, . . . , ba,).I EXAMPLE 2 The set M,(Q) of 2 X 2 matrices with entries from Q is avector space over Q. The operations area, + b, az + b,[az + b3 a4 + b4]az a4][b3 b4andba, baz[ baz bas]a1 azbLaz a4]=I EXAMPLE 3 The set Z,[x] of polynomials with coefficients from Z, is avector space over Z,, where p is a prime.I EXAMPLE 4 The set of complex numbers C = {a + bi | a, b E R} is avector space over R. The vector addition and scalar multiplication arethe usual addition and multiplication of complex numbers.The next example is a generalization of Example 4. Although itappears rather trivial, it is of the utmost importance in the theory offields.

The given set Rn=a1, a2, a3,.....,an|ai∈ℝ is a vector space over ℝ. The operations are, Vector…

Answered: I EXAMPLE 1 The set R" = {(a,, a, ...,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Verify that each of the sets in Examples 1– 4 satisfies the axioms for
a vector space. Find a basis for each of the vector spaces in
Examples 1–4.

I EXAMPLE 1 The set R" = {(a,, a, ..., a,) | a; E R} is a vector space
over R. Here the operations are the obvious ones:
%3D
(a,, a, ..., a) + (b,, b» . .. , b,) = (a, + b,, a, + b» . .., a, + b,)
1'
and
b(a,, az, . .. , a) = (ba,, ba, . . . , ba,).
I EXAMPLE 2 The set M,(Q) of 2 X 2 matrices with entries from Q is a
vector space over Q. The operations are
a, + b, az + b,
[az + b3 a4 + b4]
az a4]
[b3 b4
and
ba, baz
[ baz bas]
a1 az
b
Laz a4]
=
I EXAMPLE 3 The set Z,[x] of polynomials with coefficients from Z, is a
vector space over Z,, where p is a prime.
I EXAMPLE 4 The set of complex numbers C = {a + bi | a, b E R} is a
vector space over R. The vector addition and scalar multiplication are
the usual addition and multiplication of complex numbers.
The next example is a generalization of Example 4. Although it
appears rather trivial, it is of the utmost importance in the theory of
fields.

Expert Solution

Step 1

The given set $R^{n} = \{(a_{1}, a_{2}, a_{3}, . . . . ., a_{n}) | a_{i} \in ℝ\}$ is a vector space over $ℝ$ .

The operations are,

Vector addition: $(a_{1}, a_{2}, a_{3}, . . . . ., a_{n}) + (b_{1}, b_{2}, b_{3}, . . . . ., b_{n}) = (a_{1} + b_{1}, a_{2} + b_{2}, a_{3} + b_{3}, . . . . ., a_{n} + b_{n})$

Scalar multiplication: $b (a_{1}, a_{2}, a_{3}, . . . . ., a_{n}) = (b a_{1}, b a_{2}, b a_{3}, . . . . ., b a_{n})$ .

We have to verify that the set satisfies the vector space axioms and find its basis.

Note:

Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved, then please post only that question.

Step 2

Vector space axioms:

The set $V$ is a vector space over a field $F$ , then $V$ satisfies the following conditions:

Additive axioms:

For every $x, y, z \in V$

i) $x + y = y + x$

ii) $(x + y) + z = x + (y + z)$

iii) $0 + x = x + 0$

iv) $(- x) + x = x + (- x) = 0$

Multiplicative axioms:

For every $x \in V$ and $c, d \in F$

v) $0 . x = 0$

vi) $1 . x = x$

vii) $(c d) x = c (d x)$

Distributive axioms:

For every $x, y \in V$ and $c, d \in F$

viii) $c (x + y) = c x + c y$

ix) $(c + d) x = c x + d x$ .

Step 3

The given set $V = R^{n} = \{(a_{1}, a_{2}, a_{3}, . . . . ., a_{n}) | a_{i} \in ℝ\}$ is a vector space over $ℝ$ .

Additive axioms:

For every $x = (x_{1}, x_{2}, x_{3}, . . . ., x_{n}), y = (y_{1}, y_{2}, y_{3}, . . . ., y_{n}), z = (z_{1}, z_{2}, z_{3}, . . . ., z_{n}) \in V$

$\begin{array}{rcl} x + y & = & (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) + (y_{1}, y_{2}, y_{3}, . . . ., y_{n}) \\ = & (x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3}, . . . ., x_{n} + y_{n}) \\ = & (y_{1} + x_{1}, y_{2} + x_{2}, y_{3} + x_{3}, . . . ., y_{n} + x_{n}) \\ = & (y_{1}, y_{2}, y_{3}, . . . ., y_{n}) + (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) \\ x + y & = & y + x \end{array}$

ii)

$\begin{array}{rcl} (x + y) + z & = & ((x_{1}, x_{2}, x_{3}, . . . ., x_{n}) + (y_{1}, y_{2}, y_{3}, . . . ., y_{n})) + (z_{1}, z_{2}, z_{3}, . . . ., z_{n}) \\ = & (x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3}, . . . ., x_{n} + y_{n}) + (z_{1}, z_{2}, z_{3}, . . . ., z_{n}) \\ = & (x_{1} + y_{1} + z_{1}, x_{2} + y_{2} + z_{2}, x_{3} + y_{3} + z_{3}, . . . ., x_{n} + y_{n} + z_{n}) \\ = & (x_{1} + (y_{1} + z_{1}), x_{2} + (y_{2} + z_{2}), x_{3} + (y_{3} + z_{3}), . . . ., x_{n} + (y_{n} + z_{n})) \\ = & (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) + (y_{1} + z_{1}, y_{2} + z_{2}, y_{3} + z_{3}, . . . ., y_{n} + z_{n}) \\ (x + y) + z & = & x + (y + z) \end{array}$

iii)

$\begin{array}{rcl} 0 + x & = & (0, 0, 0, . . . . ., 0) + (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) \\ = & (0 + x_{1}, 0 + x_{2}, 0 + x_{3}, . . . ., 0 + x_{n}) \\ = & (x_{1} + 0, x_{2} + 0, x_{3} + 0, . . . ., x_{n} + 0) \\ = & (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) + (0, 0, 0, . . . . ., 0) \\ 0 + x & = & x + 0 \end{array}$

iv)

$\begin{array}{rcl} (- x) + x & = & (- x_{1}, - x_{2}, - x_{3}, . . . ., - x_{n}) + (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) \\ = & (- x_{1} + x_{1}, - x_{2} + x_{2}, - x_{3} + x_{3}, . . . ., - x_{n} + x_{n}) \\ = & (0, 0, 0, . . . ., 0) \\ (- x) + x & = & 0 \end{array}$

And,

$\begin{array}{rcl} x + (- x) & = & (x_{1}, x_{2}, x_{3}, . . . ., x_{n}) + (- x_{1}, - x_{2}, - x_{3}, . . . ., - x_{n}) \\ = & (x_{1} - x_{1}, x_{2} - x_{2}, x_{3} - x_{3}, . . . ., x_{n} - x_{n}) \\ = & (0, 0, 0, . . . ., 0) \\ x + (- x) & = & 0 \end{array}$

We get,

$(- x) + x = x + (- x) = 0$ .

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