AATA Subspaces

Section 20.2 Subspaces

Just as groups have subgroups and rings have subrings, vector spaces also have substructures. Let

V

be a vector space over a field

F,

and

W

a subset of

V .

Then

W

is a subspace of

V

if it is closed under vector addition and scalar multiplication; that is, if

u, v \in W

and

α \in F,

it will always be the case that

u + v

and

α v

are also in

W .

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Example 20.6.

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Let

W

be the subspace of

R^{3}

defined by

W = {(x_{1}, 2 x_{1} + x_{2}, x_{1} - x_{2}) : x_{1}, x_{2} \in R} .

We claim that

W

is a subspace of

R^{3} .

Since

\begin{aligned} α (x_{1}, 2 x_{1} + x_{2}, x_{1} - x_{2}) & = (α x_{1}, α (2 x_{1} + x_{2}), α (x_{1} - x_{2})) \\ = (α x_{1}, 2 (α x_{1}) + α x_{2}, α x_{1} - α x_{2}), \end{aligned}

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W

is closed under scalar multiplication. To show that

W

is closed under vector addition, let

u = (x_{1}, 2 x_{1} + x_{2}, x_{1} - x_{2})

and

v = (y_{1}, 2 y_{1} + y_{2}, y_{1} - y_{2})

be vectors in

W .

Then

u + v = (x_{1} + y_{1}, 2 (x_{1} + y_{1}) + (x_{2} + y_{2}), (x_{1} + y_{1}) - (x_{2} + y_{2})) .

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Example 20.7.

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Let

W

be the subset of polynomials of

F [x]

with no odd-power terms. If

p (x)

and

q (x)

have no odd-power terms, then neither will

p (x) + q (x) .

Also,

α p (x) \in W

for

α \in F

and

p (x) \in W .

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Let

V

be any vector space over a field

F

and suppose that

v_{1}, v_{2}, \dots, v_{n}

are vectors in

V

and

α_{1}, α_{2}, \dots, α_{n}

are scalars in

F .

Any vector

w

V

of the form

w = \sum_{i = 1}^{n} α_{i} v_{i} = α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n}

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is called a linear combination of the vectors

v_{1}, v_{2}, \dots, v_{n} .

The spanning set of vectors

v_{1}, v_{2}, \dots, v_{n}

is the set of vectors obtained from all possible linear combinations of

v_{1}, v_{2}, \dots, v_{n} .

W

is the spanning set of

v_{1}, v_{2}, \dots, v_{n},

then we say that

W

is spanned by

v_{1}, v_{2}, \dots, v_{n} .

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Proposition 20.8.

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Let

S = {v_{1}, v_{2}, \dots, v_{n}}

be vectors in a vector space

V .

Then the span of

S

is a subspace of

V .

Proof.

Let

u

and

v

be in

S .

We can write both of these vectors as linear combinations of the

v_{i}

’s:

\begin{aligned} u & = α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} \\ v & = β_{1} v_{1} + β_{2} v_{2} + \dots + β_{n} v_{n} . \end{aligned}

Then

u + v = (α_{1} + β_{1}) v_{1} + (α_{2} + β_{2}) v_{2} + \dots + (α_{n} + β_{n}) v_{n}

is a linear combination of the

v_{i}

’s. For

α \in F,

α u = (α α_{1}) v_{1} + (α α_{2}) v_{2} + \dots + (α α_{n}) v_{n}

is in the span of

S .

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