TBIL-LA Subspace Basis and Dimension (EV6)

Section 2.6 Subspace Basis and Dimension (EV6)

Learning Outcomes

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

Subsection 2.6.1 Warm Up

Activity 2.6.1.

Consider the set

S

of vectors in

R^{4}

given by

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

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(a)

Is the set

S

linearly independent or linearly dependent?

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(b)

How would you describe the subspace

span S

geometrically?

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(c)

What do the spaces

span S

and

R^{2}

have in common? In what ways do they differ?

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Subsection 2.6.2 Class Activities

Observation 2.6.2.

Recall from section Section 2.3 that a subspace of a vector space is the result of spanning a set of vectors from that vector 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 2.6.3.

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)

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)

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

Definition 2.6.4.

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 2.6.5.

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

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

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 \\ 0 \\ 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 2.6.6.

(a)

Find a basis for

span S

where

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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)

Find a basis for

span T

where

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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 2.6.7.

Even though we found different bases for them,

span S

and

span T

are exactly the same subspace of

R^{4},

since

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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 2.6.8.

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,

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{{\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}]}

are all valid bases for

R^{3},

and they all contain three vectors.

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Definition 2.6.9.

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

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{{\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}]}

contains exactly three vectors.

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Activity 2.6.10.

Consider the following subspace

W

R^{4} :

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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)

Explain and demonstrate how to find a basis of

W .

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(b)

Explain and demonstrate how to find the dimension of

W .

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Activity 2.6.11.

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

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

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Subsection 2.6.3 Individual Practice

Activity 2.6.12.

In Observation 2.6.5, we found a basis for the subspace

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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 \\ 0 \\ 1 \end{matrix}]} .

To do so, we use the results of the calculation:

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

to conclude that the set

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

the set of vectors corresponding to the pivot columns of the RREF, is a basis for

W .

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(a)

Explain why neither of the vectors

[\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}]

are elements of

W .

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(b)

Explain why this shows that, in general, when we calculate a basis for

W = span {{\vec{v}}_{1}, \dots, {\vec{v}}_{n}},

the pivot columns of

RREF [{\vec{v}}_{1} \dots {\vec{v}}_{n}]

themselves do not form a basis for

W .

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Subsection 2.6.4 Videos

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

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Subsection 2.6.5 Exercises

Exercises available at https://tbil.org/preview/linear-algebra/exercises/#/bank/EV6/.

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Subsection 2.6.6 Mathematical Writing Explorations

Exploration 2.6.13.

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 $m = n .$
If ${{\vec{v}}_{1}, {\vec{v}}_{2} \dots, {\vec{v}}_{n}}$ is linearly independent, then so is ${{\vec{v}}_{1}, {\vec{v}}_{1} + {\vec{v}}_{2}, \dots, {\vec{v}}_{1} + {\vec{v}}_{2} + \dots + {\vec{v}}_{n}} .$
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 2.6.14.

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

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Exploration 2.6.15.

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 2.6.7 Sample Problem and Solution

Sample problem Example B.1.10.

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