Short Answer

Consider a linear transformation T from $R^{2}$ to $R^{2}$ . Suppose that $\vec{v}$ and $\vec{w}$ are two arbitrary vectors in $R^{2}$ and that $\vec{x}$ is a third vector whose endpoint is on the line segment connecting the endpoints of $\vec{v}$ and $\vec{w}$ . Is the endpoint of the vector $T (\vec{x)}$ necessarily on the line segment connecting the endpoints of $T (\vec{v)}$ and $T (\vec{w)}$ ? Justify your answer.

Thus, we can say that $T (\vec{x}) = T (\vec{v}) + k (T (\vec{w}) - T \vec{(v}))$ Is on line segment.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step by step Explanation Step1: Assuming the equation

We can write $\vec{x} = \vec{v} + k (\vec{w} - \vec{v})$ where k is any scalar between 0 and 1.

Step2: Linear Transformation

A transformation from $R^{m} \to R^{n}$ is said to be linear if the following condition holds:

Identity of $R^{m}$ should be mapped to identity of role="math" localid="1659717675430" $R^{n}$ .
$T (a + b) = T (a) + T (b)$ For all $a, b \in R^{m}$ .
$T (c a) = c T (a)$ Where c is any scalar and $a \in R^{m}$ .

Step3: Solving the equations

On a similar basis we can write

$T (\vec{v} + k (\vec{w} - \vec{v}) = T (\vec{v}) + k T (\vec{w} - \vec{v}) T (\vec{v} + k (\vec{w} - \vec{v}) = T (\vec{v}) + k (T (\vec{w}) - T \vec{(v}))$

Thus, we can say that $T (\vec{x}) = T (\vec{v}) + k (T (\vec{w}) - T \vec{(v}))$ Is on line segment.

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Linear Algebra With Applications

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Step by step Explanation Step1: Assuming the equation

Step2: Linear Transformation

Step3: Solving the equations

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Linear Algebra With Applications

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Step by step Explanation Step1: Assuming the equation

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