DLA Discovery guide

Discovery guide 17.1 Discovery guide
A4 US

Discovery 17.1.

Consider the vectors

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

v_{2} = (1, 1, 2),

and

v_{3} = (1, - 1, 0) .

(a)

Do you remember what

Span

means? Explain why the vector

x = 3 v_{1} + 2 v_{2} - v_{3}

is in

Span {v_{1}, v_{2}, v_{3}} .

(b)

Actually,

v_{2}

can be expressed as a linear combination of

v_{1}

and

v_{3}

— do you see how?

Use this and the expression for

x

in Task a to express

x

as a linear combination of just

v_{1}

and

v_{3} .

(c)

Task b shows that

x

is in

Span {v_{1}, v_{3}} .

Do you think that similar calculations and the same reasoning can be carried out for every vector in

Span {v_{1}, v_{2}, v_{3}} ?

What does this say about

Span {v_{1}, v_{2}, v_{3}}

versus

Span {v_{1}, v_{3}} ?

Discovery 17.1 demonstrates a common pattern: when one of the vectors in a spanning set can be expressed as a linear combination of the others, that vector becomes redundant, and a smaller spanning set can be used in place of the original one. We’ll give this situation a name: a set of vectors is called linearly dependent if (at least) one of the vectors in the set can be written as a linear combination of other vectors in the set; otherwise the set of vectors is called linearly independent. However, it can be tedious to check each vector in a set one-by-one to see if it is a linear combination of others. Luckily, for a finite set of vectors, there is a way to check all of them all at once.

Test for Linear Dependence/Independence.

To test whether vectors

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

are linearly dependent or independent, set up the vector equation

\begin{matrix} (✶) & k_{1} v_{1} + k_{2} v_{2} + \dots + k_{m} v_{m} = 0, \end{matrix}

where the coefficients

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

are (scalar) variables.

If vector equation (✶) has a nontrivial solution in the variables $k_{1}, k_{2}, \dots, k_{m},$ then the vectors $v_{1}, v_{2}, \dots, v_{m}$ are linearly dependent.
Otherwise, if vector equation (✶) has only the trivial solution $k_{1} = 0, k_{2} = 0, \dots, k_{m} = 0,$ then the vectors $v_{1}, v_{2}, \dots, v_{m}$ are linearly independent.

Discovery 17.2.

(a)

Use the test to verify that

v_{1}, v_{2}, v_{3}

from Discovery 17.1 are linearly dependent.

(b)

Use the test to verify that

v_{1}, v_{3}

from Discovery 17.1 are linearly independent.

The next discovery activity will help you understand the Test for Linear Dependence/Independence. To keep it simple, we’ll consider just three vectors at a time.

Discovery 17.3.

(a)

Consider abstract vectors

u_{1}, u_{2}, u_{3},

and suppose the vector equation

\begin{matrix} (✶✶) & k_{1} u_{1} + k_{2} u_{2} + k_{3} u_{3} = 0 \end{matrix}

has a nontrivial solution. This means that there are values for the scalars

k_{1}, k_{2}, k_{3},

at least one of which is not zero, so that equation (✶✶) is true.

Use some algebra to manipulate equation (✶✶) to demonstrate that one of the vectors can be expressed as a linear combination of the others (and hence, by definition, the vectors

u_{1}, u_{2}, u_{3}

are linearly dependent).

(b)

Consider abstract vectors

w_{1}, w_{2}, w_{3},

and suppose the vector equation

\begin{matrix} (✶✶✶) & k_{1} w_{1} + k_{2} w_{2} + k_{3} w_{3} = 0 \end{matrix}

has only the trivial solution. We would like to see why this means that

w_{1}, w_{2}, w_{3}

are linearly independent.

Suppose they weren’t: for example, suppose

w_{3} = c_{1} w_{1} + c_{2} w_{2}

were true for some scalars

c_{1}, c_{2} .

Manipulate this expression for

w_{3}

until is says something about equation (✶✶✶). Do you see now why

w_{1}, w_{2}, w_{3}

cannot satisfy the definition of linearly dependence, and hence must be linearly independent?

Discovery 17.4.

In each of the following vector spaces, practise using the Test for Linear Dependence/Independence of the given set of vectors.

(a)

V = M_{2} (R),

S = {[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]} .

(b)

V = M_{2} (R),

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

(c)

V = P (R),

S = {1 + x, 1 + x^{2}, 2 - x + 3 x^{2}} .

Hint.

After setting up the vector equation from the test for linear dependence/independence, you are solving for the scalars

k_{1}, k_{2}, k_{3},

not for

x .

On the right-hand side, the zero represents the zero vector, which in this space is the zero polynomial. What are the coefficients on powers of

x

in the zero polynomial? The left-hand side, being equal, must have the same coefficients.

(d)

V = P (R),

S = {1, x, x^{2}, x^{3}} .

Discovery 17.5.

(a)

Do you think it’s possible to have a set of three linearly independent vectors in

R^{2} ?

Why or why not?

(b)

Do you think it’s possible to have a set of four linearly independent vectors in

R^{3} ?

Why or why not?

Discovery 17.6.

(a)

What does the definition of linear dependence say in the case of just two vectors?

(b)

If the test for linear dependence/independence is to remain true in the case of a “set” of vectors consisting of just one vector, how should we define linear dependence/independence for such a set?

Prev Top Next