Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Show that there is a nontrivial relation among the vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . . . ., {\vec{v}}_{m}$ if (and only if) at least one of the vectors is a linear combination of the other vectors
${\vec{v}}_{1}, {\vec{v}}_{2}, . . . . {\vec{v}}_{i - 1}, {\vec{v}}_{i + 1}, . . . ., {\vec{v}}_{m} .$

There is a nontrivial relation among the vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . ., {\vec{v}}_{m}$ if (and only if) at least one of the vectors ${\vec{v}}_{i}$ is a linear combination of the other vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . . . {\vec{v}}_{i - 1}, {\vec{v}}_{i + 1}, . . . ., {\vec{v}}_{m} .$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Writing the vectors v→i in a linear combination of the other vectors

Let us suppose that there is a non-trivial relation

$c_{1} v ⃗_{1} + c_{2} v ⃗_{2} + . . . + c_{i - 1} v ⃗_{i - 1} + c_{i} v ⃗_{i} + c_{i + 1} v ⃗_{i + 1} + . . . + c_{m} v ⃗_{m} = 0$ with $c_{i} \neq 0, i = 1, 2, . . ., m$

Then we have

$c_{i} v ⃗_{i} = - c_{1} v ⃗_{1} - c_{2} v ⃗_{2} - . . . - c_{i - 1} v ⃗_{i - 1} - c_{i - 1} v ⃗_{i + 1} - . . . - c_{m} v ⃗_{m}$

$\Rightarrow {\vec{v}}_{i} = - c_{i}^{- 1} c_{1} {\vec{v}}_{1} - c_{i}^{- 1} c_{2} {\vec{v}}_{2} - . . . . . - c_{i}^{- 1} c_{i - 1} {\vec{v}}_{i - 1} - c_{i}^{- 1} c_{i + 1} {\vec{v}}_{i + 1} - . . . - c_{i}^{- 1} c_{m} {\vec{v}}_{m}$

The vector is a linear combination of the other vector .

Step 2: To prove the converse part of the given statement

Conversely, let us consider that the vector is a linear combination of the other vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . . . {\vec{v}}_{i - 1}, {\vec{v}}_{i + 1}, . . . ., {\vec{v}}_{m} .$ . Then we can write

${\vec{v}}_{1} = d_{1} {\vec{v}}_{1} + d_{2} {\vec{v}}_{2} + . . . + d_{i - 1} {\vec{v}}_{i - 1} + d_{i + 1} {\vec{v}}_{i + 1} + . . . + d_{m} {\vec{v}}_{m}$

Subtracting role="math" localid="1659359195199" ${\vec{v}}_{i}$ from both sides of above equation, we get

role="math" localid="1659359178216" $0 = d_{1} {\vec{v}}_{1} + d_{2} {\vec{v}}_{2} + . . . + d_{i - 1} {\vec{v}}_{i - 1} + d_{i + 1} {\vec{v}}_{i + 1} + . . . + d_{m} {\vec{v}}_{m}$

Thus, we get a non-trivial relation among the vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . ., {\vec{v}}_{m} .$ if the vector is a linear combination of the other vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . . . {\vec{v}}_{i - 1}, {\vec{v}}_{i + 1}, . . . ., {\vec{v}}_{m} .$ .

Step 3: Final Answer

There is a nontrivial relation among the vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . ., {\vec{v}}_{m} .$ if (and only if) at least one of the vectors ${\vec{v}}_{i}$ is a linear combination of the other vectors ${\vec{v}}_{1}, {\vec{v}}_{2}, . . . . {\vec{v}}_{i - 1}, {\vec{v}}_{i + 1}, . . . ., {\vec{v}}_{m} .$ .

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: Writing the vectors v→i in a linear combination of the other vectors

Step 2: To prove the converse part of the given statement

Step 3: Final Answer

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Linear Algebra With Applications

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Short Answer

Show that there is a nontrivial relation among the vectors v→1,v→2,.....,v→m if (and only if) at least one of the vectors is a linear combination of the other vectorsv→1,v→2,....v→i-1,v→i+1,....,v→m.

Step by Step Solution

Step 1: Writing the vectors v→i in a linear combination of the other vectors

Step 2: To prove the converse part of the given statement

Step 3: Final Answer

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