DLA Discovery guide

Discovery guide 17.1 Discovery guide
A4 US

Discovery 17.1.

Consider the vectors \(\uvec{v}_1 = (1,0,1)\text{,}\) \(\uvec{v}_2 = (1,1,2)\text{,}\) and \(\uvec{v}_3 = (1,-1,0)\text{.}\)

(a)

Do you remember what \(\Span\) means? Explain why the vector

\begin{equation*} \uvec{x} = 3\uvec{v}_1 + 2\uvec{v}_2 - \uvec{v}_3 \end{equation*}

is in \(\Span \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\text{.}\)

(b)

Actually, \(\uvec{v}_2\) can be expressed as a linear combination of \(\uvec{v}_1\) and \(\uvec{v}_3\) — do you see how?

Use this and the expression for \(\uvec{x}\) in Task a to express \(\uvec{x}\) as a linear combination of just \(\uvec{v}_1\) and \(\uvec{v}_3\text{.}\)

(c)

Task b shows that \(\uvec{x}\) is in \(\Span \{\uvec{v}_1,\uvec{v}_3\}\text{.}\) Do you think that similar calculations and the same reasoning can be carried out for every vector in \(\Span \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\text{?}\)

What does this say about \(\Span \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\) versus \(\Span \{\uvec{v}_1,\uvec{v}_3\}\text{?}\)

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

\begin{equation*} \uvec{v}_1, \uvec{v}_2, \dotsc, \uvec{v}_m \end{equation*}

are linearly dependent or independent, set up the vector equation

\begin{gather} k_1\uvec{v}_1 + k_2\uvec{v}_2 + \dotsb + k_m\uvec{v}_m = \zerovec\text{,}\tag{✶} \end{gather}

where the coefficients \(k_1,k_2,\dotsc,k_m\) are (scalar) variables.

If vector equation (✶) has a nontrivial solution in the variables \(k_1,k_2,\dotsc,k_m\text{,}\) then the vectors \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_m\) are linearly dependent.
Otherwise, if vector equation (✶) has only the trivial solution \(k_1=0,k_2=0,\dotsc,k_m=0\text{,}\) then the vectors \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_m\) are linearly independent.

Discovery 17.2.

(a)

Use the test to verify that \(\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\) from Discovery 17.1 are linearly dependent.

(b)

Use the test to verify that \(\uvec{v}_1,\uvec{v}_3\) 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 \(\uvec{u}_1,\uvec{u}_2,\uvec{u}_3\text{,}\) and suppose the vector equation

\begin{gather} k_1\uvec{u}_1 + k_2\uvec{u}_2 + k_3\uvec{u}_3 = \zerovec\tag{✶✶} \end{gather}

has a nontrivial solution. This means that there are values for the scalars \(k_1,k_2,k_3\text{,}\) 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 \(\uvec{u}_1,\uvec{u}_2,\uvec{u}_3\) are linearly dependent).

(b)

Consider abstract vectors \(\uvec{w}_1,\uvec{w}_2,\uvec{w}_3\text{,}\) and suppose the vector equation

\begin{gather} k_1\uvec{w}_1 + k_2\uvec{w}_2 + k_3\uvec{w}_3 = \zerovec\tag{✶✶✶} \end{gather}

has only the trivial solution. We would like to see why this means that \(\uvec{w}_1,\uvec{w}_2,\uvec{w}_3\) are linearly independent.

Suppose they weren’t: for example, suppose \(\uvec{w}_3 = c_1\uvec{w}_1 + c_2\uvec{w}_2\) were true for some scalars \(c_1,c_2\text{.}\) Manipulate this expression for \(\uvec{w}_3\) until is says something about equation (✶✶✶). Do you see now why \(\uvec{w}_1,\uvec{w}_2,\uvec{w}_3\) 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)

\(V = \matrixring_2(\R)\text{,}\) \(S= \left\{\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}, \;\; \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix}, \;\; \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix} \; \right\} \text{.}\)

(b)

\(V = \matrixring_2(\R)\text{,}\) \(S= \left\{\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}, \;\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}, \;\; \begin{bmatrix} 3 \amp 0 \\ 0 \amp -2 \end{bmatrix} \; \right\} \text{.}\)

(c)

\(V = \poly(\R)\text{,}\) \(S = \{ 1+x, 1+x^2, 2 - x + 3x^2 \}\text{.}\)

Hint.

After setting up the vector equation from the test for linear dependence/independence, you are solving for the scalars \(k_1,k_2,k_3\text{,}\) not for \(x\text{.}\) 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)

\(V = \poly(\R)\text{,}\) \(S = \{ 1, x, x^2, x^3 \}\text{.}\)

Discovery 17.5.

(a)

Do you think it’s possible to have a set of three linearly independent vectors in \(\R^2\text{?}\) Why or why not?

(b)

Do you think it’s possible to have a set of four linearly independent vectors in \(\R^3\text{?}\) 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