DLA Discovery guide

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Discovery guide 19.1 Discovery guide

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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.

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This leads to the following definition.

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

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Discovery 19.1.

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.

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(a)

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

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(b)

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

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(c)

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

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(d)

$V =$ the space of $2\times 2$ upper triangular matrices, $S = \left\{\; \left[\begin{smallmatrix} 1 \amp 0 \\ 0 \amp 1 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 1 \\ 0 \amp 1 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 2 \\ 0 \amp 1 \end{smallmatrix}\right],\;\; \left[\begin{smallmatrix} 1 \amp 3 \\ 0 \amp 1 \end{smallmatrix}\right] \;\right\} \text{.}$

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(e)

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

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

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(f)

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

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(g)

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

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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.

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Discovery 19.2.

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.

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(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:

$\begin{equation*} (3,-3) \; = \; \underline{\hspace{0.909090909090909em}}(1,0) \; + \; \underline{\hspace{0.909090909090909em}}(1,1) \; + \; \underline{\hspace{0.909090909090909em}}(1,-1)\text{.} \end{equation*}$

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(b)

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

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(c)

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

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The next discovery activity will demonstrate that the redundancies of Discovery 19.2 cannot happen for a basis.

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Discovery 19.3.

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 $S\text{.}$

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Discovery 19.3 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:

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

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This is called the coordinate vector of

$\uvec{w}$ relative to

$S$ .

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Discovery 19.4.

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

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(a)

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

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(b)

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

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(c)

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

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(d)

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

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(e)

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

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Discovery 19.5.

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

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(a)

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(b)

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(c)

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

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(d)

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

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(e)

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

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Discovery 19.6.

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 19.5.b.

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(a)

In Task 19.5.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 $\rmatrixOfplain{\uvec{v}}{S} = (-1,2,0,3)\text{.}$

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(b)

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

Using your vectors $\uvec{v}$ from Task a, and $\uvec{w}$ from Task 19.5.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.

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(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 19.6.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.

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(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 19.5.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 19.6.b. Surprised?