AATA Linear Independence

Section 20.3 Linear Independence

Let

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

be a set of vectors in a vector space

V .

If there exist scalars

α_{1}, α_{2} \dots α_{n} \in F

such that not all of the

α_{i}

’s are zero and

α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} = 0,

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then

S

is said to be linearly dependent. If the set

S

is not linearly dependent, then it is said to be linearly independent. More specifically,

S

is a linearly independent set if

α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} = 0

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implies that

α_{1} = α_{2} = \dots = α_{n} = 0

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for any set of scalars

{α_{1}, α_{2} \dots α_{n}} .

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Proposition 20.9.

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Let

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

be a set of linearly independent vectors in a vector space. Suppose that

v = α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} = β_{1} v_{1} + β_{2} v_{2} + \dots + β_{n} v_{n} .

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Then

α_{1} = β_{1}, α_{2} = β_{2}, \dots, α_{n} = β_{n} .

Proof.

v = α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} = β_{1} v_{1} + β_{2} v_{2} + \dots + β_{n} v_{n},

then

(α_{1} - β_{1}) v_{1} + (α_{2} - β_{2}) v_{2} + \dots + (α_{n} - β_{n}) v_{n} = 0 .

Since

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

are linearly independent,

α_{i} - β_{i} = 0

for

i = 1, \dots, n .

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The definition of linear dependence makes more sense if we consider the following proposition.

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Proposition 20.10.

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A set

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

of vectors in a vector space

V

is linearly dependent if and only if one of the

v_{i}

’s is a linear combination of the rest.

Proof.

Suppose that

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

is a set of linearly dependent vectors. Then there exist scalars

α_{1}, \dots, α_{n}

such that

α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} = 0,

with at least one of the

α_{i}

’s not equal to zero. Suppose that

α_{k} \neq 0 .

Then

v_{k} = - \frac{α_{1}}{α_{k}} v_{1} - \dots - \frac{α_{k - 1}}{α_{k}} v_{k - 1} - \frac{α_{k + 1}}{α_{k}} v_{k + 1} - \dots - \frac{α_{n}}{α_{k}} v_{n} .

Conversely, suppose that

v_{k} = β_{1} v_{1} + \dots + β_{k - 1} v_{k - 1} + β_{k + 1} v_{k + 1} + \dots + β_{n} v_{n} .

Then

β_{1} v_{1} + \dots + β_{k - 1} v_{k - 1} - v_{k} + β_{k + 1} v_{k + 1} + \dots + β_{n} v_{n} = 0 .

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The following proposition is a consequence of the fact that any system of homogeneous linear equations with more unknowns than equations will have a nontrivial solution. We leave the details of the proof for the end-of-chapter exercises.

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Proposition 20.11.

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Suppose that a vector space

V

is spanned by

n

vectors. If

m > n,

then any set of

m

vectors in

V

must be linearly dependent.

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A set

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

of vectors in a vector space

V

is called a basis for

V

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

is a linearly independent set that spans

V .

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Example 20.12.

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The vectors

e_{1} = (1, 0, 0),

e_{2} = (0, 1, 0),

and

e_{3} = (0, 0, 1)

form a basis for

R^{3} .

The set certainly spans

R^{3},

since any arbitrary vector

(x_{1}, x_{2}, x_{3})

R^{3}

can be written as

x_{1} e_{1} + x_{2} e_{2} + x_{3} e_{3} .

Also, none of the vectors

e_{1}, e_{2}, e_{3}

can be written as a linear combination of the other two; hence, they are linearly independent. The vectors

e_{1}, e_{2}, e_{3}

are not the only basis of

R^{3} :

the set

{(3, 2, 1), (3, 2, 0), (1, 1, 1)}

is also a basis for

R^{3} .

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Example 20.13.

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Let

Q (\sqrt{2}) = {a + b \sqrt{2} : a, b \in Q} .

The sets

{1, \sqrt{2}}

and

{1 + \sqrt{2}, 1 - \sqrt{2}}

are both bases of

Q (\sqrt{2}) .

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From the last two examples it should be clear that a given vector space has several bases. In fact, there are an infinite number of bases for both of these examples. In general, there is no unique basis for a vector space. However, every basis of

R^{3}

consists of exactly three vectors, and every basis of

Q (\sqrt{2})

consists of exactly two vectors. This is a consequence of the next proposition.

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Proposition 20.14.

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Let

{e_{1}, e_{2}, \dots, e_{m}}

and

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

be two bases for a vector space

V .

Then

m = n .

Proof.

Since

{e_{1}, e_{2}, \dots, e_{m}}

is a basis, it is a linearly independent set. By Proposition 20.11,

n \leq m .

Similarly,

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

is a linearly independent set, and the last proposition implies that

m \leq n .

Consequently,

m = n .

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

is a basis for a vector space

V,

then we say that the dimension of

V

n

and we write

\dim V = n .

We will leave the proof of the following theorem as an exercise.

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Theorem 20.15.

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Let

V

be a vector space of dimension

n .

If $S = {v_{1}, \dots, v_{n}}$ is a set of linearly independent vectors for $V,$ then $S$ is a basis for $V .$
If $S = {v_{1}, \dots, v_{n}}$ spans $V,$ then $S$ is a basis for $V .$
If $S = {v_{1}, \dots, v_{k}}$ is a set of linearly independent vectors for $V$ with $k < n,$ then there exist vectors $v_{k + 1}, \dots, v_{n}$ such that

${v_{1}, \dots, v_{k}, v_{k + 1}, \dots, v_{n}}$

is a basis for $V .$

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