TBIL-LA Linear Transformations (AT1)

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Section 3.1 Linear Transformations (AT1)

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Learning Outcomes

Determine if a map between vector spaces of polynomials is linear or not.

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Subsection 3.1.1 Class Activities

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Definition 3.1.1.

A linear transformation (also called a linear map) is a map between vector spaces that preserves the vector space operations. More precisely, if $V$ and $W$ are vector spaces, a map $T : V \to W$ is called a linear transformation if

$T (\vec{v} + \vec{w}) = T (\vec{v}) + T (\vec{w})$ for any $\vec{v}, \vec{w} \in V,$ and
$T (c \vec{v}) = c T (\vec{v})$ for any $c \in R,$ and $\vec{v} \in V .$

In other words, a map is linear when vector space operations can be applied before or after the transformation without affecting the result.

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Definition 3.1.2.

Given a linear transformation $T : V \to W,$ $V$ is called the domain of $T$ and $W$ is called the co-domain of $T .$

Figure 23. A linear transformation with a domain of $R^{3}$ and a co-domain of $R^{2}$

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Example 3.1.3.

Let $T : R^{3} \to R^{2}$ be given by

T ([\begin{matrix} x \\ y \\ z \end{matrix}]) = [\begin{matrix} x - z \\ 3 y \end{matrix}] .

To show that $T$ is a linear transformation, we must verify that $T (\vec{v} + \vec{w}) = T (\vec{v}) + T (\vec{w})$ by computing

T ([\begin{matrix} x \\ y \\ z \end{matrix}] + [\begin{matrix} u \\ v \\ w \end{matrix}]) = T ([\begin{matrix} x + u \\ y + v \\ z + w \end{matrix}]) = [\begin{matrix} (x + u) - (z + w) \\ 3 (y + v) \end{matrix}]

and

T ([\begin{matrix} x \\ y \\ z \end{matrix}]) + T ([\begin{matrix} u \\ v \\ w \end{matrix}]) = [\begin{matrix} x - z \\ 3 y \end{matrix}] + [\begin{matrix} u - w \\ 3 v \end{matrix}] = [\begin{matrix} (x + u) - (z + w) \\ 3 (y + v) \end{matrix}],

and we must verify that $T (c \vec{v}) = c T (\vec{v})$ by computing

T (c [\begin{matrix} x \\ y \\ z \end{matrix}]) = T ([\begin{matrix} c x \\ c y \\ c z \end{matrix}]) = [\begin{matrix} c x - c z \\ 3 c y \end{matrix}] and c T ([\begin{matrix} x \\ y \\ z \end{matrix}]) = c [\begin{matrix} x - z \\ 3 y \end{matrix}] = [\begin{matrix} c x - c z \\ 3 c y \end{matrix}] .

Therefore $T$ is a linear transformation.

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Example 3.1.4.

Let $S : R^{2} \to R^{4}$ be given by

S ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} x + y \\ x^{2} \\ y + 3 \\ y - 2^{x} \end{matrix}]

To show that $S$ is not linear, we only need to find one counterexample.

S ([\begin{matrix} 0 \\ 1 \end{matrix}] + [\begin{matrix} 2 \\ 3 \end{matrix}]) = S ([\begin{matrix} 2 \\ 4 \end{matrix}]) = [\begin{matrix} 6 \\ 4 \\ 7 \\ 0 \end{matrix}]

S ([\begin{matrix} 0 \\ 1 \end{matrix}]) + S ([\begin{matrix} 2 \\ 3 \end{matrix}]) = [\begin{matrix} 1 \\ 0 \\ 4 \\ 0 \end{matrix}] + [\begin{matrix} 5 \\ 4 \\ 6 \\ - 1 \end{matrix}] = [\begin{matrix} 6 \\ 4 \\ 10 \\ - 1 \end{matrix}]

Since the resulting vectors are different, $S$ is not a linear transformation.

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Fact 3.1.5.

A map between Euclidean spaces $T : R^{n} \to R^{m}$ is linear exactly when every component of the output is a linear combination of the variables of $R^{n} .$

For example, the following map is definitely linear because $x - z$ and $3 y$ are linear combinations of $x, y, z :$

T ([\begin{matrix} x \\ y \\ z \end{matrix}]) = [\begin{matrix} x - z \\ 3 y \end{matrix}] = [\begin{matrix} 1 x + 0 y - 1 z \\ 0 x + 3 y + 0 z \end{matrix}] .

But the map below is not linear because $x^{2},$ $y + 3,$ and $y - 2^{x}$ are not linear combinations (even though $x + y$ is):

S ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} x + y \\ x^{2} \\ y + 3 \\ y - 2^{x} \end{matrix}] .

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Activity 3.1.6.

Let $D : P \to P$ be the derivative map defined by $D (f (x)) = f^{'} (x)$ for each polynomial $f \in P .$ We recall from calculus that

D (f (x) + g (x)) = f^{'} (x) + g^{'} (x),

and

D (c f (x)) = c f^{'} (x) .

Which of the following can we conclude from these calculus rules?

$P$ is not a vector space
$D$ is a linear map
$D$ is not a linear map

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Activity 3.1.7.

Let the polynomial maps $S : P_{4} \to P_{3}$ and $T : P_{4} \to P_{3}$ be defined by

S (f (x)) = 2 f^{'} (x) - f^{″} (x) T (f (x)) = f^{'} (x) + x^{3} .

Compute $S (x^{4} + x),$ $S (x^{4}) + S (x),$ $T (x^{4} + x),$ and $T (x^{4}) + T (x) .$ Based on these computations, can you conclude that either $S$ or $T$ is definitely not a linear transformation?

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Fact 3.1.8.

If $L : V \to W$ is a linear transformation, then $L (\vec{z}) = L (0 \vec{v}) = 0 L (\vec{v}) = \vec{z}$ where $\vec{z}$ is the additive identity of the vector spaces $V, W .$

Put another way, an easy way to prove that a map like $T (f (x)) = f^{'} (x) + x^{3}$ can not be linear is to check that

T (0) = \frac{d}{d x} [0] + x^{3} = 0 + x^{3} = x^{3} \neq 0.

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Observation 3.1.9.

Showing $T : V \to W$ is not a linear transformation can be done by finding an example for any one of the following.

Show $T (\vec{z}) \neq \vec{z}$ (where $\vec{z}$ is the additive identity of $V$ and $W$ ).
Find $\vec{v}, \vec{w} \in V$ such that $T (\vec{v} + \vec{w}) \neq T (\vec{v}) + T (\vec{w}) .$
Find $\vec{v} \in V$ and $c \in R$ such that $T (c \vec{v}) \neq c T (\vec{v}) .$

Otherwise, $T$ can be shown to be linear by proving the following in general.

For all $\vec{v}, \vec{w} \in V,$ $T (\vec{v} + \vec{w}) = T (\vec{v}) + T (\vec{w}) .$
For all $\vec{v} \in V$ and $c \in R,$ $T (c \vec{v}) = c T (\vec{v}) .$

Note the similarities between this process and showing that a subset of a vector space is or is not a subspace.

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Activity 3.1.10.

Continue to consider $S : P_{4} \to P_{3}$ defined by

S (f (x)) = 2 f^{'} (x) - f^{″} (x) .

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(a)

Verify that

S (f (x) + g (x)) = 2 f^{'} (x) + 2 g^{'} (x) - f^{″} (x) - g^{″} (x)

is equal to $S (f (x)) + S (g (x))$ for all polynomials $f, g \in P_{4} .$

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(b)

Verify that $S (c f (x))$ is equal to $c S (f (x))$ for all real numbers $c$ and polynomials $f \in P_{4} .$

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(c)

Is $S$ linear?

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Activity 3.1.11.

Let polynomial maps $S : P \to P$ and $T : P \to P$ be defined by

S (f (x)) = (f (x))^{2} T (f (x)) = 3 x f (x^{2})

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(a)

Note that $S (0) = 0$ and $T (0) = 0 .$ So instead, show that $S (x + 1) \neq S (x) + S (1)$ to verify that $S$ is not linear.

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(b)

Prove that $T$ is linear by verifying that $T (f (x) + g (x)) = T (f (x)) + T (g (x))$ and $T (c f (x)) = c T (f (x)) .$

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Subsection 3.1.2 Videos

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Figure 24. Video: Showing a transformation is linear

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Figure 25. Video: Showing a transformation is not linear

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Exploration 3.1.12.

If $V, W$ are vectors spaces, with associated zero vectors ${\vec{0}}_{V}$ and ${\vec{0}}_{W},$ and $T : V \to W$ is a linear transformation, does $T ({\vec{0}}_{V}) = {\vec{0}}_{W} ?$ Prove this is true, or find a counterexample.

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Exploration 3.1.13.

Assume $f : V \to W$ is a linear transformation between vector spaces. Let $\vec{v} \in V$ with additive inverse ${\vec{v}}^{- 1} .$ Prove that $f ({\vec{v}}^{- 1}) = [f (\vec{v})]^{- 1} .$

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Subsection 3.1.6 Sample Problem and Solution

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Sample problem Example B.1.14.

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