A list of $v$ of one vector $v in V$ is linear independent if and only if $v neq 0$
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Why a list of $v$ of one vector $v in V$ is linear independent if and only if $v neq 0$?
To my understanding, if the only choice of $a_1, ...a_m in mathbb{F}$ that makes $a_1v_1 + ...+ a_mv_m$ equal $0$ is $a_1=...=a_m=0$, then the list $v1,...,v_m$ of vector $V$ is called linear independence.
If so, is it because zero vector itself also counts as one way to makes $a_1v_1 + ...+ a_mv_m$ equal $0$?
linear-algebra
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
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add a comment |
$begingroup$
Why a list of $v$ of one vector $v in V$ is linear independent if and only if $v neq 0$?
To my understanding, if the only choice of $a_1, ...a_m in mathbb{F}$ that makes $a_1v_1 + ...+ a_mv_m$ equal $0$ is $a_1=...=a_m=0$, then the list $v1,...,v_m$ of vector $V$ is called linear independence.
If so, is it because zero vector itself also counts as one way to makes $a_1v_1 + ...+ a_mv_m$ equal $0$?
linear-algebra
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
$endgroup$
add a comment |
$begingroup$
Why a list of $v$ of one vector $v in V$ is linear independent if and only if $v neq 0$?
To my understanding, if the only choice of $a_1, ...a_m in mathbb{F}$ that makes $a_1v_1 + ...+ a_mv_m$ equal $0$ is $a_1=...=a_m=0$, then the list $v1,...,v_m$ of vector $V$ is called linear independence.
If so, is it because zero vector itself also counts as one way to makes $a_1v_1 + ...+ a_mv_m$ equal $0$?
linear-algebra
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
$endgroup$
Why a list of $v$ of one vector $v in V$ is linear independent if and only if $v neq 0$?
To my understanding, if the only choice of $a_1, ...a_m in mathbb{F}$ that makes $a_1v_1 + ...+ a_mv_m$ equal $0$ is $a_1=...=a_m=0$, then the list $v1,...,v_m$ of vector $V$ is called linear independence.
If so, is it because zero vector itself also counts as one way to makes $a_1v_1 + ...+ a_mv_m$ equal $0$?
linear-algebra
linear-algebra
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
asked Dec 29 '18 at 23:36
JOHN JOHN
4589
4589
add a comment |
add a comment |
3 Answers
3
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Take $vin V$. If $ainmathbb F$, when do we have $av=0$?
- If $vneq0$, that will happen only when $a=0$. Therefore, ${v}$ is linearly independent.
- If $v=0$ then, for instance, $1v=0$. Therefore ${v}$ is linearly dependent.
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
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If $v=0$, then $1cdot v=0$ and since $1neq 0$, $(v)$ is linearly dependent.
If $(v)$ is linearly dependent, then there exists $lambdainmathbb{R}^*$ with $lambda v=0$ and multiplying by $1/lambda$ leads to $v=0$.
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
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If $vneq 0$, then if there is $alphain k$ ($k$ is the ground field) such that $alpha v=0$, then if $alpha=0$ we are done, if not: multiplying by $alpha^{-1}$ we get that $0=alpha^{-1}cdot 0=alpha^{-1}cdotleft(alpha vright)=left(alpha^{-1}alpharight)v=1cdot v=vneq 0$. So that, $leftlbrace vrightrbrace$ is linealy independent.
On the other hand, if $v=0$, taking $alpha=1$, we have that $alpha v=0$, with $alphaneq 0$, then $leftlbrace vrightrbrace$ is linealy dependent.
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $vin V$. If $ainmathbb F$, when do we have $av=0$?
- If $vneq0$, that will happen only when $a=0$. Therefore, ${v}$ is linearly independent.
- If $v=0$ then, for instance, $1v=0$. Therefore ${v}$ is linearly dependent.
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
add a comment |
$begingroup$
Take $vin V$. If $ainmathbb F$, when do we have $av=0$?
- If $vneq0$, that will happen only when $a=0$. Therefore, ${v}$ is linearly independent.
- If $v=0$ then, for instance, $1v=0$. Therefore ${v}$ is linearly dependent.
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
add a comment |
$begingroup$
Take $vin V$. If $ainmathbb F$, when do we have $av=0$?
- If $vneq0$, that will happen only when $a=0$. Therefore, ${v}$ is linearly independent.
- If $v=0$ then, for instance, $1v=0$. Therefore ${v}$ is linearly dependent.
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
Take $vin V$. If $ainmathbb F$, when do we have $av=0$?
- If $vneq0$, that will happen only when $a=0$. Therefore, ${v}$ is linearly independent.
- If $v=0$ then, for instance, $1v=0$. Therefore ${v}$ is linearly dependent.
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
answered Dec 29 '18 at 23:40
José Carlos SantosJosé Carlos Santos
176k24137247
176k24137247
add a comment |
add a comment |
$begingroup$
If $v=0$, then $1cdot v=0$ and since $1neq 0$, $(v)$ is linearly dependent.
If $(v)$ is linearly dependent, then there exists $lambdainmathbb{R}^*$ with $lambda v=0$ and multiplying by $1/lambda$ leads to $v=0$.
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
$endgroup$
add a comment |
$begingroup$
If $v=0$, then $1cdot v=0$ and since $1neq 0$, $(v)$ is linearly dependent.
If $(v)$ is linearly dependent, then there exists $lambdainmathbb{R}^*$ with $lambda v=0$ and multiplying by $1/lambda$ leads to $v=0$.
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
$endgroup$
add a comment |
$begingroup$
If $v=0$, then $1cdot v=0$ and since $1neq 0$, $(v)$ is linearly dependent.
If $(v)$ is linearly dependent, then there exists $lambdainmathbb{R}^*$ with $lambda v=0$ and multiplying by $1/lambda$ leads to $v=0$.
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
$endgroup$
If $v=0$, then $1cdot v=0$ and since $1neq 0$, $(v)$ is linearly dependent.
If $(v)$ is linearly dependent, then there exists $lambdainmathbb{R}^*$ with $lambda v=0$ and multiplying by $1/lambda$ leads to $v=0$.
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
answered Dec 29 '18 at 23:39
C. FalconC. Falcon
15.3k41951
15.3k41951
add a comment |
add a comment |
$begingroup$
If $vneq 0$, then if there is $alphain k$ ($k$ is the ground field) such that $alpha v=0$, then if $alpha=0$ we are done, if not: multiplying by $alpha^{-1}$ we get that $0=alpha^{-1}cdot 0=alpha^{-1}cdotleft(alpha vright)=left(alpha^{-1}alpharight)v=1cdot v=vneq 0$. So that, $leftlbrace vrightrbrace$ is linealy independent.
On the other hand, if $v=0$, taking $alpha=1$, we have that $alpha v=0$, with $alphaneq 0$, then $leftlbrace vrightrbrace$ is linealy dependent.
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
If $vneq 0$, then if there is $alphain k$ ($k$ is the ground field) such that $alpha v=0$, then if $alpha=0$ we are done, if not: multiplying by $alpha^{-1}$ we get that $0=alpha^{-1}cdot 0=alpha^{-1}cdotleft(alpha vright)=left(alpha^{-1}alpharight)v=1cdot v=vneq 0$. So that, $leftlbrace vrightrbrace$ is linealy independent.
On the other hand, if $v=0$, taking $alpha=1$, we have that $alpha v=0$, with $alphaneq 0$, then $leftlbrace vrightrbrace$ is linealy dependent.
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
If $vneq 0$, then if there is $alphain k$ ($k$ is the ground field) such that $alpha v=0$, then if $alpha=0$ we are done, if not: multiplying by $alpha^{-1}$ we get that $0=alpha^{-1}cdot 0=alpha^{-1}cdotleft(alpha vright)=left(alpha^{-1}alpharight)v=1cdot v=vneq 0$. So that, $leftlbrace vrightrbrace$ is linealy independent.
On the other hand, if $v=0$, taking $alpha=1$, we have that $alpha v=0$, with $alphaneq 0$, then $leftlbrace vrightrbrace$ is linealy dependent.
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
If $vneq 0$, then if there is $alphain k$ ($k$ is the ground field) such that $alpha v=0$, then if $alpha=0$ we are done, if not: multiplying by $alpha^{-1}$ we get that $0=alpha^{-1}cdot 0=alpha^{-1}cdotleft(alpha vright)=left(alpha^{-1}alpharight)v=1cdot v=vneq 0$. So that, $leftlbrace vrightrbrace$ is linealy independent.
On the other hand, if $v=0$, taking $alpha=1$, we have that $alpha v=0$, with $alphaneq 0$, then $leftlbrace vrightrbrace$ is linealy dependent.
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 29 '18 at 23:42
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
add a comment |
add a comment |
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