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Q43E

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Found in: Page 132

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Question: Consider three linearly independent vectors $\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{1}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{2}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{3}}}$ in ${{\mathbf{ℝ}}}^{{\mathbf{3}}}$. Are the vectors $\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{1}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{1}}}{\mathbf{+}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{2}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{1}}}{\mathbf{+}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{2}}}{\mathbf{+}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{3}}}$ linearly independent as well? How can you tell?

Yes, the vectors $\stackrel{\to }{{\mathrm{v}}_{1}},\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}},\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}}+\stackrel{\to }{{\mathrm{v}}_{3}}$ are linearly independent.

See the step by step solution

## Step 1: Definition of Linearly independent

Linearly independent is defined as the property of a set having no linear combination of its elements equal to zero when the coefficients are taken from a given set unless the coefficient of each element is zero.

## Step 2: Prove the vectors are linearly independent

$\stackrel{\to }{{\mathrm{v}}_{1}},\stackrel{\to }{{\mathrm{v}}_{2}},\stackrel{\to }{{\mathrm{v}}_{3}}$ are linearly independent vectors.

Since $\stackrel{\to }{{\mathrm{v}}_{1}},\stackrel{\to }{{\mathrm{v}}_{2}},\stackrel{\to }{{\mathrm{v}}_{3}}$ we know that

$\left({x}_{1},{x}_{2},{x}_{3}\right)\stackrel{\to }{{\mathrm{v}}_{1}},\left({x}_{2},{x}_{3}\right)\stackrel{\to }{{\mathrm{v}}_{2}},{x}_{3}\stackrel{\to }{{\mathrm{v}}_{3}}=0⇒\left({x}_{1},{x}_{2},{x}_{3}\right)=\left({x}_{2},{x}_{3}\right)={x}_{3}=0$

From here

${x}_{3}=0\phantom{\rule{0ex}{0ex}}{x}_{1}+{x}_{2}=0\to {x}_{2}=0\phantom{\rule{0ex}{0ex}}{x}_{1},{x}_{2},{x}_{3}=0\to {x}_{1}=0$

The equation is true only in case those are 0.

role="math" localid="1659358612562" ${x}_{1}\stackrel{\to }{{\mathrm{v}}_{1}}+{x}_{2}\left(\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}}\right)+{x}_{3}\left(\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}}+\stackrel{\to }{{\mathrm{v}}_{3}}\right)=0$

From here it is clear that the vectors are linearly independent.

## Step 3: The final answer

Yes, the vectors $\stackrel{\to }{{\mathrm{v}}_{1}},\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}},\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}}+\stackrel{\to }{{\mathrm{v}}_{3}}$ are linearly independent.