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Expert-verified Found in: Page 132 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider some linearly independent vectors in and a vector ${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{\mathbf{1}\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{\mathbf{2}\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{\mathbf{m}\mathbf{,}}$ in ${{\mathbf{ℝ}}}^{{\mathbf{n}}}{\mathbf{}}{\mathbit{a}}{\mathbit{n}}{\mathbit{d}}{\mathbf{}}{\mathbit{a}}{\mathbf{}}{\mathbit{v}}{\mathbit{e}}{\mathbit{c}}{\mathbit{t}}{\mathbit{o}}{\mathbit{r}}\stackrel{\mathbf{\to }}{\mathbf{v}}{\mathbf{}}{\mathbit{i}}{\mathbit{n}}{\mathbf{}}{{\mathbf{ℝ}}}^{{\mathbf{n}}}$ that is not contained in the span ${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{v}}_{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{m}}}$ of . Are the vectors ${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{\mathbf{1}\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{\mathbf{2}\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{m}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{}}}$ necessarily linearly independent?

If there are some linearly independent vectors in and a vector ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},..{\stackrel{\to }{v}}_{m},$ in ${\mathrm{ℝ}}^{n}andavector\stackrel{\to }{v}in{\mathrm{ℝ}}^{n}$that is not contained in the span of ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{m}$ then the vectors are necessarily linearly independent.

See the step by step solution

## Step 1:   Linear Relations

Consider the vectors ${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{m}}}$ in . An equation of the form

${{\mathbit{a}}}_{{\mathbf{1}}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbit{a}}}_{{\mathbf{2}}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbit{a}}}_{{\mathbf{m}}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{m}}}{\mathbf{=}}{\mathbf{0}}$ (1)

is called a (linear) relation among the vectors ${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{3}}}$. There is always the trivial relation, with ${{\mathbit{a}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{=}}{{\mathbit{a}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{=}}{{\mathbit{a}}}_{{\mathbf{m}}}$. Nontrivial relations (where at least one coefficient ${{\mathbit{a}}}_{{\mathbf{i}}}$ is nonzero) may or may not exist among the vectors ${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{m}}}{\mathbf{.}}$ .

## Step 2: To prove the vectors v→1,v→2,...v→m,v→  are linearly independent

Since the vector $\stackrel{\to }{v}$ in ${\mathrm{ℝ}}^{n}$is not contained in the span of ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{m},$. Therefore

$\stackrel{\to }{v}\ne {a}_{1}{\stackrel{\to }{v}}_{1},+{a}_{2}{\stackrel{\to }{v}}_{2}+...+{a}_{m}{\stackrel{\to }{v}}_{m}$, for ${a}_{i}\ne 0$ . (2)

Now multiply the equation (2) by role="math" localid="1659416449027" $b\in \mathrm{ℝ}andb\ne 0$ from left ,then we have

$b\stackrel{\to }{v}\ne b{a}_{1}{\stackrel{\to }{v}}_{1}+b{a}_{2}{\stackrel{\to }{v}}_{2}+...+b{a}_{m}{\stackrel{\to }{v}}_{m}$ (3)

Now subtract both sides by $b\stackrel{\to }{v},$ , we get

role="math" localid="1659416709972" $0\ne b{a}_{1}{\stackrel{\to }{v}}_{1}+b{a}_{2}{\stackrel{\to }{v}}_{2}+...b{a}_{m}{\stackrel{\to }{v}}_{m}-b\stackrel{\to }{v}$ (4)

Since $b\ne 0⇒b{a}_{i}\ne 0$, so the right hand side of equation (4) will be zero if and only if $b{a}_{i}=0$.

Hence, the vectors ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{m}{\stackrel{\to }{v}}_{}$are linearly independent.

If there are some linearly independent vectors in ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{m}in{\mathrm{ℝ}}^{n}$ and a vector ${\mathrm{ℝ}}^{n}$ in that is not contained in the span ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{m}$ of then the vectors ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{m},\stackrel{\to }{v}$ are necessarily linearly independent. 