DLA Discovery guide

Discovery guide 18.1 Discovery guide
A4 US

Suppose \(V\) is a vector space and \(S\) is a finite spanning set for \(V\) (i.e. \(V = \Span S\)). In the previous chapter, we saw that if \(S\) is linearly dependent, then (at least) one vector can be removed from \(S\text{,}\) and the resulting smaller set will still be a spanning set. You can imagine repeating this process until finally you are left with a spanning set that is linearly independent.

This leads to the following definition.

basis for a vector space: a linearly independent spanning set for the space

Discovery 18.1.

In each of the following, determine whether \(S\) is a basis for \(V\text{.}\) If it is not a basis, make sure you know which property \(S\) violates, independence or spanning.

(a)

\(V = \R^3\text{,}\) \(S = \{(1,0,0),(1,1,0),(1,1,1),(0,0,2)\}\text{.}\)

(b)

\(V = \R^3\text{,}\) \(S = \{(1,0,0),(1,1,0),(0,0,2)\}\text{.}\)

(c)

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

(d)

\(V =\) the space of \(2\times 2\) upper triangular matrices,

\(S = \left\{\; \left[\begin{smallmatrix} 1 \amp 0 \\ 0 \amp 1 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 1 \\ 0 \amp 1 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 2 \\ 0 \amp 1 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 3 \\ 0 \amp 1 \end{smallmatrix}\right] \;\right\} \text{.}\)

(e)

\(V =\) the space of \(3\times 3\) lower triangular matrices,

\(S = \left\{\; \left[\begin{smallmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{smallmatrix}\right] \;\right\} \text{.}\)

(f)

\(V = \poly_3(\R)\text{,}\) the space of all polynomials of degree \(3\) or less, \(S = \{1,x,x^2\}\text{.}\)

(g)

\(V = \poly_3(\R)\text{,}\) \(S = \{1,x,x^2,x^3\}\text{.}\)

As discussed in the introduction to this discovery guide above, a spanning set that is not linearly independent contains redundant information in the form of vectors that are not actually needed to form a spanning set. This redundancy manifests itself in other ways, as the next discovery activity will demonstrate.

Discovery 18.2.

Consider the set \(S = \{(1,0),(1,1),(1,-1)\}\) of vectors in \(\R^2\text{.}\) This set spans \(\R^2\) but is not linearly independent.

(a)

Since \(S\) spans \(\R^2\text{,}\) it is possible to express vector \((3,-3)\) as a linear combination of the vectors in \(S\text{.}\)

Demonstrate a way to do this:

\begin{equation*} (3,-3) \; = \; \fillinmath{XX}(1,0) \; + \; \fillinmath{XX}(1,1) \; + \; \fillinmath{XX}(1,-1)\text{.} \end{equation*}

(b)

Here is the redundant part. Demonstrate a different way to express \((3,-3)\) as a linear combination of the vectors in \(S\text{:}\)

\begin{equation*} (3,-3) \; = \; \fillinmath{XX}(1,0) \; + \; \fillinmath{XX}(1,1) \; + \; \fillinmath{XX}(1,-1)\text{.} \end{equation*}

(c)

How many different ways do you think there are to do this?

The next discovery activity will demonstrate that the redundancies of Discovery 18.2 cannot happen for a basis.

Discovery 18.3.

Suppose \(V\) is a vector space, \(S = \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\) is a basis for \(V\text{,}\) and \(\uvec{w}\) is a vector in \(V\text{.}\)

Since \(S\) is a spanning set, there is a way to express \(\uvec{w}\) as linear combinations of the vectors in \(S\text{:}\)

\begin{equation*} \uvec{w} = a_1\uvec{v}_1+a_2\uvec{v}_2+a_3\uvec{v}_3\text{.} \end{equation*}

Suppose there were a different such expression:

\begin{equation*} \uvec{w} = b_1\uvec{v}_1+b_2\uvec{v}_2+b_3\uvec{v}_3\text{.} \end{equation*}

Use the vector identity

\begin{equation*} \uvec{w}-\uvec{w} = \zerovec \end{equation*}

and the two different expressions for \(\uvec{w}\) above to show that having these two different expressions violates the linear independence of \(S\text{.}\)

Discovery 18.3 shows that when we have a basis \(S = \{\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_n\}\) for a vector space \(V\text{,}\) each vector in \(V\) has one unique expression as a linear combination of the vectors in \(S\text{.}\) For \(\uvec{w} = c_1\uvec{v}_1 + c_2\uvec{v}_2 + \dotsb + c_n\uvec{v}_n\text{,}\) the coefficients \(c_1,c_2,\dotsc,c_n\) are called the coordinates of \(\uvec{w}\) relative to \(S\). Since these coordinates consist of \(n\) coefficients, we sometimes relate \(\uvec{w}\) to a vector in \(\R^n\) by collecting its coordinates into an \(n\)-tuple:

\begin{equation*} \rmatrixOfplain{\uvec{w}}{S} = (c_1,c_2,\dotsc,c_n)\text{.} \end{equation*}

This is called the coordinate vector of \(\uvec{w}\) relative to \(S\).

Discovery 18.4.

In each of the following, determine the coordinate vector of \(\uvec{w}\) relative to the provided basis \(S\) for \(V\text{.}\)

(a)

\(V = \matrixring_2(\R)\text{,}\) \(S = \left\{\; \left[\begin{smallmatrix} 1 \amp 0 \\ 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 1 \\ 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \\ 1 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \\ 0 \amp 1 \end{smallmatrix}\right] \;\right\} \text{,}\) \(\uvec{w} = \left[\begin{smallmatrix} -1 \amp 5 \\ 3 \amp -2 \end{smallmatrix}\right]\text{.}\)

(b)

\(V = \matrixring_2(\R)\text{,}\) \(S = \left\{\; \left[\begin{smallmatrix} 1 \amp 0 \\ 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 1 \\ 0 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \\ 1 \amp 0 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 0 \amp 0 \\ 1 \amp 1 \end{smallmatrix}\right] \;\right\} \text{,}\) \(\uvec{w} = \left[\begin{smallmatrix} -1 \amp 5 \\ 3 \amp -2 \end{smallmatrix}\right]\text{.}\)

(c)

\(V = \poly_3(\R)\text{,}\) \(S = \{1,x,x^2,x^3\}\text{,}\) \(\uvec{w} = 3 + 4x - 5x^3\text{.}\)

(d)

\(V = \R^3\text{,}\) \(S = \{(-1,0,1),(0,2,0),(1,1,0)\}\text{,}\) \(\uvec{w} = (1,1,1)\text{.}\)

(e)

\(V = \R^3\text{,}\) \(S = \{(1,0,0),(0,1,0),(0,0,1)\}\text{,}\) \(\uvec{w} = (-2,3,-5)\text{.}\)

Discovery 18.5.

In each of the following, determine which vector \(\uvec{w}\) in \(V\) has the given coordinate vector \(\rmatrixOfplain{\uvec{w}}{S}\text{.}\)

(a)

(b)

(c)

\(V = P_3\text{,}\) \(S = \{1,x,x^2,x^3\}\text{,}\) \(\rmatrixOfplain{\uvec{w}}{S} = (-3,1,0,3)\text{.}\)

(d)

\(V = \R^3\text{,}\) \(S = \{(-1,0,1),(0,2,0),(1,1,0)\}\text{,}\) \(\rmatrixOfplain{\uvec{w}}{S} = (1,1,1)\text{.}\)

(e)

\(V = \R^3\text{,}\) \(S = \{(1,0,0),(0,1,0),(0,0,1)\}\text{,}\) \(\rmatrixOfplain{\uvec{w}}{S} = (-2,3,-5)\text{.}\)

Discovery 18.6.

Coordinate vectors let us transfer vector algebra in a space \(V\) to the familiar space \(\R^n\text{.}\)

For example, consider the basis

\begin{equation*} S = \left\{\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix},\;\; \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix},\;\; \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix},\;\; \begin{bmatrix} 0 \amp 0 \\ 1 \amp 1 \end{bmatrix} \;\right\} \end{equation*}

for the space \(\matrixring_2(\R)\) from Task 18.5.b.

(a)

In Task 18.5.b you have already determined the vector \(\uvec{w}\) in \(\matrixring_2(\R)\) that has coordinate vector \(\rmatrixOfplain{\uvec{w}}{S} = (3,-5,1,1)\text{.}\) Now do the same to determine the vector \(\uvec{v}\) in \(\matrixring_2(\R)\) that has coordinate vector \(\rmatrixOfplain{\uvec{v}}{S} = (-1,2,0,3)\text{.}\)

(b)

Do some algebra in \(\matrixring_2(\R)\text{:}\)

Using your vectors \(\uvec{v}\) from Task a, and \(\uvec{w}\) from Task 18.5.b compute the linear combination \(2 \uvec{v} + \uvec{w}\text{.}\)

Note: Vectors \(\uvec{v}\) and \(\uvec{w}\) “live” in the space \(\matrixring_2(\R)\text{,}\) so your computation in this task should involve \(2 \times 2\) matrices, and should also result in a \(2 \times 2\) matrix.

(c)

Do the same algebra in \(\R^4\text{:}\)

Compute \(2 \rmatrixOfplain{\uvec{v}}{S} + \rmatrixOfplain{\uvec{w}}{S}\text{,}\) using the coordinate vectors \(\rmatrixOfplain{\uvec{v}}{S} \) and \(\rmatrixOfplain{\uvec{w}}{S}\) provided to you in Task 18.6.a.

Note: These coordinate vectors “live” in the space \(\R^4\text{,}\) so your computation in this task should involve four-dimensional vectors, and should also result in a four-dimensional vector.

(d)

Compare your results:

Consider your four-dimensional result vector from Task c as a coordinate vector for some vector in \(\matrixring_2(\R)\) relative to \(S\text{.}\) Similarly to your computations in Task 18.5.b and Task a, determine the matrix in \(\matrixring_2(\R)\) that has coordinate vector equal to your result vector from Task c. Then compare with your result matrix from Task 18.6.b. Surprised?

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