TBIL-LA Subspace Basis and Dimension (EV6)

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Section 2.6 Subspace Basis and Dimension (EV6)

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Learning Outcomes

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Compute a basis for the subspace spanned by a given set of Euclidean vectors, and determine the dimension of the subspace.

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Subsection 2.6.1 Class Activities

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Observation 2.6.1.

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Recall from section Section 2.3 that a subspace of a vector space is the result of spanning a set of vectors from that space.

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Recall also that a linearly dependent set contains “redundant” vectors. For example, only two of the three vectors in Figure 14 are needed to span the planar subspace.

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Activity 2.6.2.

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Consider the subspace of

R^{4}

given by

W = span {[\begin{matrix} 2 \\ 3 \\ 0 \\ 1 \end{matrix}], [\begin{matrix} 2 \\ 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ - 3 \\ 2 \\ - 3 \end{matrix}], [\begin{matrix} 1 \\ 5 \\ - 1 \\ 0 \end{matrix}]} .

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(a)

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Mark the column of

RREF [\begin{array}{cccc} 2 & 2 & 2 & 1 \\ 3 & 0 & - 3 & 5 \\ 0 & 1 & 2 & - 1 \\ 1 & - 1 & - 3 & 0 \end{array}]

that shows that

W

’s spanning set is linearly dependent.

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(b)

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What would be the result of removing the vector that gave us this column?

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The set still spans $W,$ and remains linearly dependent.
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The set still spans $W,$ but is now also linearly independent.
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The set no longer spans $W,$ and remains linearly dependent.
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The set no longer spans $W,$ but is now linearly independent.


    
        
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rref([2,2,2,1; 3,0,-3,5; 0,1,2,-1; 1,-1,-3,0])

    
    
    
    
        
            
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Definition 2.6.3.

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Let

W

be a subspace of a vector space. A basis for

W

is a linearly independent set of vectors that spans

W

(but not necessarily the entire vector space).

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Observation 2.6.4.

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So given a set

S = {{\vec{v}}_{1}, \dots, {\vec{v}}_{m}},

to compute a basis for the subspace

span S,

simply remove the vectors corresponding to the non-pivot columns of

RREF [{\vec{v}}_{1} \dots {\vec{v}}_{m}] .

For example, since

RREF [\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 2 & 4 & - 2 & 2 \\ 3 & 6 & - 2 & 1 \end{array}] = [\begin{array}{cccc} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}]

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the subspace

W = span {[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}], [\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}], [\begin{matrix} 0 \\ - 2 \\ - 2 \end{matrix}], [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}]}

has

{[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}], [\begin{matrix} 0 \\ - 2 \\ - 2 \end{matrix}]}

as a basis.

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Activity 2.6.5.

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(a)

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Find a basis for

span S

where

S = {[\begin{matrix} 2 \\ 3 \\ 0 \\ 1 \end{matrix}], [\begin{matrix} 2 \\ 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ - 3 \\ 2 \\ - 3 \end{matrix}], [\begin{matrix} 1 \\ 5 \\ - 1 \\ 0 \end{matrix}]} .

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(b)

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Find a basis for

span T

where

T = {[\begin{matrix} 2 \\ 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ - 3 \\ 2 \\ - 3 \end{matrix}], [\begin{matrix} 1 \\ 5 \\ - 1 \\ 0 \end{matrix}], [\begin{matrix} 2 \\ 3 \\ 0 \\ 1 \end{matrix}]} .

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Observation 2.6.6.

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Even though we found different bases for them,

span S

and

span T

are exactly the same subspace of

R^{4},

since

S = {[\begin{matrix} 2 \\ 3 \\ 0 \\ 1 \end{matrix}], [\begin{matrix} 2 \\ 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ - 3 \\ 2 \\ - 3 \end{matrix}], [\begin{matrix} 1 \\ 5 \\ - 1 \\ 0 \end{matrix}]} = {[\begin{matrix} 2 \\ 0 \\ 1 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ - 3 \\ 2 \\ - 3 \end{matrix}], [\begin{matrix} 1 \\ 5 \\ - 1 \\ 0 \end{matrix}], [\begin{matrix} 2 \\ 3 \\ 0 \\ 1 \end{matrix}]} = T .

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Thus the basis for a subspace is not unique in general.

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Fact 2.6.7.

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Any non-trivial real vector space has infinitely-many different bases, but all the bases for a given vector space are exactly the same size.

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For example,

{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}} and {[\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}], [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]} and {[\begin{matrix} 1 \\ 0 \\ - 3 \end{matrix}], [\begin{matrix} 2 \\ - 2 \\ 1 \end{matrix}], [\begin{matrix} 3 \\ - 2 \\ 5 \end{matrix}]}

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are all valid bases for

R^{3},

and they all contain three vectors.

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Definition 2.6.8.

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The dimension of a vector space or subspace is equal to the size of any basis for the vector space.

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As you’d expect,

R^{n}

has dimension

n .

For example,

R^{3}

has dimension

3

because any basis for

R^{3}

such as

{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}} and {[\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}], [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]} and {[\begin{matrix} 1 \\ 0 \\ - 3 \end{matrix}], [\begin{matrix} 2 \\ - 2 \\ 1 \end{matrix}], [\begin{matrix} 3 \\ - 2 \\ 5 \end{matrix}]}

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contains exactly three vectors.

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Activity 2.6.9.

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Consider the following subspace

W

R^{4} :

W = span {[\begin{matrix} 1 \\ 0 \\ 0 \\ - 1 \end{matrix}], [\begin{matrix} - 2 \\ 0 \\ 0 \\ 2 \end{matrix}], [\begin{matrix} - 3 \\ 1 \\ - 5 \\ 5 \end{matrix}], [\begin{matrix} 12 \\ - 3 \\ 15 \\ - 18 \end{matrix}]} .

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(a)

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Explain and demonstrate how to find a basis of

W .

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(b)

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Explain and demonstrate how to find the dimension of

W .

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Activity 2.6.10.

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The dimension of a subspace may be found by doing what with an appropriate RREF matrix?

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Count the rows.
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Count the non-pivot columns.
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Count the pivots.
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Add the number of pivot rows and pivot columns.

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Subsection 2.6.2 Videos

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Figure 17. Video: Finding a basis of a subspace and computing the dimension of a subspace

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Exploration 2.6.11.

Prove each of the following statements is true.

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If ${{\vec{b}}_{1}, {\vec{b}}_{2}, \dots, {\vec{b}}_{m}}$ and ${{\vec{c}}_{1}, {\vec{c}}_{2}, \dots, {\vec{c}}_{n}}$ are each a basis for a vector space $V,$ then $m = n .$
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If ${{\vec{v}}_{1}, {\vec{v}}_{2} \dots, {\vec{v}}_{n}}$ is linearly independent, then so is ${{\vec{v}}_{1}, {\vec{v}}_{1} + {\vec{v}}_{2}, \dots, {\vec{v}}_{1} + {\vec{v}}_{2} + \dots + {\vec{v}}_{n}} .$
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Let $V$ be a vector space of dimension $n,$ and $\vec{v} \in V .$ Then there exists a basis for $V$ which contains $\vec{v} .$

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Exploration 2.6.12.

Suppose we have the set of all function

f : S \to R .

We claim that this is a vector space under the usual operation of function addition and scalar multiplication. What is the dimension of this space for each choice of

S

below:

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$S = {1}$
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$S = {1, 2}$
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$S = {1, 2, \dots, n}$
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$S = R$

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Exploration 2.6.13.

Suppose you have the vector space

V = {(\begin{matrix} x \\ y \\ z \end{matrix}) \in R^{3} : x + y + z = 1}

with the operations

(\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) \oplus (\begin{matrix} x_{2} \\ y_{2} \\ z_{2} \end{matrix}) = (\begin{matrix} x_{1} + x_{2} - 1 \\ y_{1} + y_{2} \\ z_{1} + z_{2} \end{matrix}) and α ⊙ (\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}) = (\begin{matrix} α x_{1} - α + 1 \\ α y_{1} \\ α z_{1} \end{matrix}) .

Find a basis for

V

and determine it’s dimension.

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Subsection 2.6.6 Sample Problem and Solution

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Sample problem Example B.1.10.

Linear Algebra for Team-Based Inquiry Learning: 2023 Edition

Search Results:

Section 2.6 Subspace Basis and Dimension (EV6)

Learning Outcomes

Subsection 2.6.1 Class Activities

Observation 2.6.1.

Activity 2.6.2.

(a)

(b)

Definition 2.6.3.

Observation 2.6.4.

Activity 2.6.5.

(a)

(b)

Observation 2.6.6.

Fact 2.6.7.

Definition 2.6.8.

Activity 2.6.9.

(a)

(b)

Activity 2.6.10.

Subsection 2.6.2 Videos

Subsection 2.6.3 Slideshow

Exercises 2.6.4 Exercises

Subsection 2.6.5 Mathematical Writing Explorations

Exploration 2.6.11.

Exploration 2.6.12.

Exploration 2.6.13.

Subsection 2.6.6 Sample Problem and Solution