TBIL-LA Identifying a Basis (EV5)

🔗

Activity 2.5.2.

🔗

Let’s review some of the terminology we’ve been dealing with...

🔗

(a)

🔗

If every vector in a vector space can be constructed as one or more linear combination of vectors in a set

S,

we can say...

the set $S$ spans the vector space.
the set $S$ fails to span the vector space.
the set $S$ is linearly independent.
the set $S$ is linearly dependent.

🔗

(b)

🔗

If the zero vector

\vec{0}

can be constructed as a unique linear combination of vectors in a set

S

(the combination multiplying every vector by the scalar value

0

), we can say...

the set $S$ spans the vector space.
the set $S$ fails to span the vector space.
the set $S$ is linearly independent.
the set $S$ is linearly dependent.

🔗

(c)

🔗

If every vector of a vector space can either be constructed as a unique linear combination of vectors in a set

S,

or not at all, we can say...

the set $S$ spans the vector space.
the set $S$ fails to span the vector space.
the set $S$ is linearly independent.
the set $S$ is linearly dependent.

🔗

Definition 2.5.3.

🔗

A basis of a vector space

V

is a set of vectors

S

contained in

V

for which

Every vector in the vector space can be expressed as a linear combination of the vectors in $S .$
For each vector $\vec{v}$ in the vector space, there is only one way to write it as a linear combination of the vectors in $S .$

🔗

These two properties may be expressed more succintly as the statement "Every vector in

V

can be expressed uniquely as a linear combination of the vectors in

S

🔗

Observation 2.5.4.

🔗

In terms of a vector equation, a set

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

is a basis of a vector space if the vector equation

x_{1} \vec{v_{1}} + \dots + x_{n} \vec{v_{n}} = \vec{w}

🔗

has a unique solution for every vector

\vec{w}

in the vector space.

🔗

Put another way, a basis may be thought of as a minimal set of “building blocks” that can be used to construct any other vector of the vector space.

🔗

Activity 2.5.5.

🔗

Let

S

be a basis (Definition 2.5.3) for a vector space. Then...

the set $S$ must both span the vector space and be linearly independent.
the set $S$ must span the vector space but could be linearly dependent.
the set $S$ must be linearly independent but could fail to span the vector space.
the set $S$ could fail to span the vector space and could be linearly dependent.

🔗

Activity 2.5.6.

🔗

The vectors

\begin{aligned} \hat{i} & = (1, 0, 0) = [\begin{array}{c} 1 \\ 0 \\ 0 \end{array}] & \hat{j} & = (0, 1, 0) = [\begin{array}{c} 0 \\ 1 \\ 0 \end{array}] & \hat{k} & = (0, 0, 1) = [\begin{array}{c} 0 \\ 0 \\ 1 \end{array}] \end{aligned}

🔗

form a basis

{\hat{i}, \hat{j}, \hat{k}}

used frequently in multivariable calculus.

🔗

Find the unique linear combination of these vectors

? \hat{i} + ? \hat{j} + ? \hat{k}

🔗

that equals the vector

(3, - 2, 4) = [\begin{matrix} 3 \\ - 2 \\ 4 \end{matrix}]

🔗

x y z

space.

🔗

Definition 2.5.7.

🔗

The standard basis of

R^{n}

is the set

{{\vec{e}}_{1}, \dots, {\vec{e}}_{n}}

where

\begin{aligned} {\vec{e}}_{1} & = [\begin{array}{c} 1 \\ 0 \\ 0 \\ ⋮ \\ 0 \\ 0 \end{array}] & {\vec{e}}_{2} & = [\begin{array}{c} 0 \\ 1 \\ 0 \\ ⋮ \\ 0 \\ 0 \end{array}] & \dots & {\vec{e}}_{n} = [\begin{array}{c} 0 \\ 0 \\ 0 \\ ⋮ \\ 0 \\ 1 \end{array}] . \end{aligned}

🔗

In particular, the standard basis for

R^{3}

{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}} = {\hat{i}, \hat{j}, \hat{k}} .

🔗

Activity 2.5.8.

🔗

Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for

R^{4} .

🔗

(a)

🔗

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

A basis, because it both spans $R^{4}$ and is linearly independent.
Not a basis, because while it spans $R^{4},$ it is linearly dependent.
Not a basis, because while it is linearly independent, it fails to span $R^{4} .$
Not a basis, because not only does it fail to span $R^{4},$ it’s also linearly dependent.

🔗

(b)

🔗

{[\begin{matrix} 2 \\ 3 \\ 0 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ 0 \\ 0 \\ 3 \end{matrix}], [\begin{matrix} 4 \\ 3 \\ 0 \\ 2 \end{matrix}], [\begin{matrix} - 3 \\ 0 \\ 1 \\ 3 \end{matrix}]}

A basis, because it both spans $R^{4}$ and is linearly independent.
Not a basis, because while it spans $R^{4},$ it is linearly dependent.
Not a basis, because while it is linearly independent, it fails to span $R^{4} .$
Not a basis, because not only does it fail to span $R^{4},$ it’s also linearly dependent.

🔗

(c)

🔗

{[\begin{matrix} 2 \\ 3 \\ 0 \\ - 1 \end{matrix}], [\begin{matrix} 2 \\ 0 \\ 0 \\ 3 \end{matrix}], [\begin{matrix} 3 \\ 13 \\ 7 \\ 16 \end{matrix}], [\begin{matrix} - 1 \\ 10 \\ 7 \\ 14 \end{matrix}], [\begin{matrix} 4 \\ 3 \\ 0 \\ 2 \end{matrix}]}

A basis, because it both spans $R^{4}$ and is linearly independent.
Not a basis, because while it spans $R^{4},$ it is linearly dependent.
Not a basis, because while it is linearly independent, it fails to span $R^{4} .$
Not a basis, because not only does it fail to span $R^{4},$ it’s also linearly dependent.

🔗

(d)

🔗

{[\begin{matrix} 2 \\ 3 \\ 0 \\ - 1 \end{matrix}], [\begin{matrix} 4 \\ 3 \\ 0 \\ 2 \end{matrix}], [\begin{matrix} - 3 \\ 0 \\ 1 \\ 3 \end{matrix}], [\begin{matrix} 3 \\ 6 \\ 1 \\ 5 \end{matrix}]}

A basis, because it both spans $R^{4}$ and is linearly independent.
Not a basis, because while it spans $R^{4},$ it is linearly dependent.
Not a basis, because while it is linearly independent, it fails to span $R^{4} .$
Not a basis, because not only does it fail to span $R^{4},$ it’s also linearly dependent.

🔗

(e)

🔗

{[\begin{matrix} 5 \\ 3 \\ 0 \\ - 1 \end{matrix}], [\begin{matrix} - 2 \\ 1 \\ 0 \\ 3 \end{matrix}], [\begin{matrix} 4 \\ 5 \\ 1 \\ 3 \end{matrix}]}

A basis, because it both spans $R^{4}$ and is linearly independent.
Not a basis, because while it spans $R^{4},$ it is linearly dependent.
Not a basis, because while it is linearly independent, it fails to span $R^{4} .$
Not a basis, because not only does it fail to span $R^{4},$ it’s also linearly dependent.

🔗

Activity 2.5.9.

🔗

{{\vec{v}}_{1}, {\vec{v}}_{2}, {\vec{v}}_{3}, {\vec{v}}_{4}}

is a basis for

R^{4},

that means

RREF [{\vec{v}}_{1} {\vec{v}}_{2} {\vec{v}}_{3} {\vec{v}}_{4}]

has a pivot in every row (because it spans), and has a pivot in every column (because it’s linearly independent).

🔗

What is

RREF [{\vec{v}}_{1} {\vec{v}}_{2} {\vec{v}}_{3} {\vec{v}}_{4}] ?

RREF [{\vec{v}}_{1} {\vec{v}}_{2} {\vec{v}}_{3} {\vec{v}}_{4}] = [\begin{array}{cccc} ? & ? & ? & ? \\ ? & ? & ? & ? \\ ? & ? & ? & ? \\ ? & ? & ? & ? \end{array}]

🔗

Fact 2.5.10.

🔗

The set

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

is a basis for

R^{n}

if and only if

m = n

and

RREF [{\vec{v}}_{1} \dots {\vec{v}}_{n}] = [\begin{array}{cccc} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 1 \end{array}] .

🔗

That is, a basis for

R^{n}

must have exactly

n

vectors and its square matrix must row-reduce to the so-called identity matrix containing all zeros except for a downward diagonal of ones. (We will learn where the identity matrix gets its name in a later module.)

🔗

Subsection 2.5.2 Videos

🔗

Figure 16. Video: Verifying that a set of vectors is a basis of a vector space

🔗

Exercises 2.5.3 Exercises

🔗

Exercises available at https://teambasedinquirylearning.github.io/linear-algebra/2024e/exercises/#/bank/EV5/.

🔗

Subsection 2.5.4 Mathematical Writing Explorations

🔗

Exploration 2.5.11.

What is a basis for $M_{2, 2} ?$
What about $M_{3, 3} ?$
Could we write each of these in a way that looks like the standard basis vectors in $R^{m}$ for some $m ?$ Make a conjecture about the relationship between these spaces of matrices and standard Eulidean space.

🔗

Exploration 2.5.12.

Recall our earlier definition of symmetric matrices. Find a basis for each of the following:

The space of $2 \times 2$ symmetric matrices.
The space of $3 \times 3$ symmetric matrices.
The space of $n \times n$ symmetric matrices.

🔗

Exploration 2.5.13.

Must a basis for the space

P_{2},

the space of all quadratic polynomials, contain a polynomial of each degree less than or equal to 2? Generalize your result to polynomials of arbitrary degree.

🔗

Subsection 2.5.5 Sample Problem and Solution

🔗

Sample problem Example B.1.9.

Prev Top Next