Is linear dependence dependent on field?
$begingroup$
Given the vector space $mathbb{C}^3$ and three vectors $x_1 = (1, 0, 0)$, $x_2 = (0, 1, 0)$, $x_3 = (0, 0, 1)$.
Is it correct that these three vectors are linearly independent if $mathbb{C}^3$ is defined over the field $mathbb{R}$, while they are linearly dependent if the field is $mathbb{C}$?
I'm using the following definiton for linear dependence (Halmos, Finite-Dimensional Vector Spaces, 2e):
A finite set of vectors ${x_i}$ is linearly dependent if there
exists a corresponding set ${alpha_i}$ of scalars, not all zero,
such that
$$ tag{1}label{eqn_li}sum_i alpha_i x_i = 0, $$
and the reason I'm asking is that $eqref{eqn_li}$
is satisfied for the scalars $alpha_1 = i$ , $alpha_2 = 0$, $alpha_3 = 0$, hence there is a set of scalars ${alpha_i}$, not all zeros, such that $eqref{eqn_li}$ holds.
linear-algebra vector-spaces
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
$endgroup$
add a comment |
$begingroup$
Given the vector space $mathbb{C}^3$ and three vectors $x_1 = (1, 0, 0)$, $x_2 = (0, 1, 0)$, $x_3 = (0, 0, 1)$.
Is it correct that these three vectors are linearly independent if $mathbb{C}^3$ is defined over the field $mathbb{R}$, while they are linearly dependent if the field is $mathbb{C}$?
I'm using the following definiton for linear dependence (Halmos, Finite-Dimensional Vector Spaces, 2e):
A finite set of vectors ${x_i}$ is linearly dependent if there
exists a corresponding set ${alpha_i}$ of scalars, not all zero,
such that
$$ tag{1}label{eqn_li}sum_i alpha_i x_i = 0, $$
and the reason I'm asking is that $eqref{eqn_li}$
is satisfied for the scalars $alpha_1 = i$ , $alpha_2 = 0$, $alpha_3 = 0$, hence there is a set of scalars ${alpha_i}$, not all zeros, such that $eqref{eqn_li}$ holds.
linear-algebra vector-spaces
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
$endgroup$
$begingroup$
"...for the scalars..." . What scalars??
$endgroup$
– DonAntonio
Jan 12 at 17:44
5
$begingroup$
The given set of scalars doesn't satisfy (1).
$endgroup$
– Thomas Shelby
Jan 12 at 17:45
1
$begingroup$
Hint: $sumalpha_jx_j=(alpha_1,alpha_2,alpha_3)$.
$endgroup$
– David C. Ullrich
Jan 12 at 18:03
add a comment |
$begingroup$
Given the vector space $mathbb{C}^3$ and three vectors $x_1 = (1, 0, 0)$, $x_2 = (0, 1, 0)$, $x_3 = (0, 0, 1)$.
Is it correct that these three vectors are linearly independent if $mathbb{C}^3$ is defined over the field $mathbb{R}$, while they are linearly dependent if the field is $mathbb{C}$?
I'm using the following definiton for linear dependence (Halmos, Finite-Dimensional Vector Spaces, 2e):
A finite set of vectors ${x_i}$ is linearly dependent if there
exists a corresponding set ${alpha_i}$ of scalars, not all zero,
such that
$$ tag{1}label{eqn_li}sum_i alpha_i x_i = 0, $$
and the reason I'm asking is that $eqref{eqn_li}$
is satisfied for the scalars $alpha_1 = i$ , $alpha_2 = 0$, $alpha_3 = 0$, hence there is a set of scalars ${alpha_i}$, not all zeros, such that $eqref{eqn_li}$ holds.
linear-algebra vector-spaces
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
$endgroup$
Given the vector space $mathbb{C}^3$ and three vectors $x_1 = (1, 0, 0)$, $x_2 = (0, 1, 0)$, $x_3 = (0, 0, 1)$.
Is it correct that these three vectors are linearly independent if $mathbb{C}^3$ is defined over the field $mathbb{R}$, while they are linearly dependent if the field is $mathbb{C}$?
I'm using the following definiton for linear dependence (Halmos, Finite-Dimensional Vector Spaces, 2e):
A finite set of vectors ${x_i}$ is linearly dependent if there
exists a corresponding set ${alpha_i}$ of scalars, not all zero,
such that
$$ tag{1}label{eqn_li}sum_i alpha_i x_i = 0, $$
and the reason I'm asking is that $eqref{eqn_li}$
is satisfied for the scalars $alpha_1 = i$ , $alpha_2 = 0$, $alpha_3 = 0$, hence there is a set of scalars ${alpha_i}$, not all zeros, such that $eqref{eqn_li}$ holds.
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
edited Jan 12 at 20:53
José Carlos Santos
157k22126227
157k22126227
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
asked Jan 12 at 17:41
Max HerrmannMax Herrmann
672416
672416
$begingroup$
"...for the scalars..." . What scalars??
$endgroup$
– DonAntonio
Jan 12 at 17:44
5
$begingroup$
The given set of scalars doesn't satisfy (1).
$endgroup$
– Thomas Shelby
Jan 12 at 17:45
1
$begingroup$
Hint: $sumalpha_jx_j=(alpha_1,alpha_2,alpha_3)$.
$endgroup$
– David C. Ullrich
Jan 12 at 18:03
add a comment |
$begingroup$
"...for the scalars..." . What scalars??
$endgroup$
– DonAntonio
Jan 12 at 17:44
5
$begingroup$
The given set of scalars doesn't satisfy (1).
$endgroup$
– Thomas Shelby
Jan 12 at 17:45
1
$begingroup$
Hint: $sumalpha_jx_j=(alpha_1,alpha_2,alpha_3)$.
$endgroup$
– David C. Ullrich
Jan 12 at 18:03
$begingroup$
"...for the scalars..." . What scalars??
$endgroup$
– DonAntonio
Jan 12 at 17:44
$begingroup$
"...for the scalars..." . What scalars??
$endgroup$
– DonAntonio
Jan 12 at 17:44
5
5
$begingroup$
The given set of scalars doesn't satisfy (1).
$endgroup$
– Thomas Shelby
Jan 12 at 17:45
$begingroup$
The given set of scalars doesn't satisfy (1).
$endgroup$
– Thomas Shelby
Jan 12 at 17:45
1
1
$begingroup$
Hint: $sumalpha_jx_j=(alpha_1,alpha_2,alpha_3)$.
$endgroup$
– David C. Ullrich
Jan 12 at 18:03
$begingroup$
Hint: $sumalpha_jx_j=(alpha_1,alpha_2,alpha_3)$.
$endgroup$
– David C. Ullrich
Jan 12 at 18:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Those three vectors are linearly independent both over $mathbb C$ and over $mathbb R$.
However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $mathbb C$ and linearly independent over $mathbb R$.
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Your assumption that $ix_1+0x_2+0x_3=0$ is wrong.
Your three vectors are linearly independent, no matter if we view $Bbb C^3$ as three-dimensional space over $Bbb C$, or six-dimensional space over $Bbb R$ (or e.g., infinite-dimensional space over $Bbb Q$) in the apparent way.
However, $(1,0,0)$ and $(i,0,0)$ are $Bbb R$-linearly independant and $Bbb C$-linearly dependant.
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Those three vectors are linearly independent both over $mathbb C$ and over $mathbb R$.
However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $mathbb C$ and linearly independent over $mathbb R$.
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Those three vectors are linearly independent both over $mathbb C$ and over $mathbb R$.
However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $mathbb C$ and linearly independent over $mathbb R$.
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Those three vectors are linearly independent both over $mathbb C$ and over $mathbb R$.
However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $mathbb C$ and linearly independent over $mathbb R$.
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
Those three vectors are linearly independent both over $mathbb C$ and over $mathbb R$.
However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $mathbb C$ and linearly independent over $mathbb R$.
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 12 at 17:44
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
add a comment |
add a comment |
$begingroup$
Your assumption that $ix_1+0x_2+0x_3=0$ is wrong.
Your three vectors are linearly independent, no matter if we view $Bbb C^3$ as three-dimensional space over $Bbb C$, or six-dimensional space over $Bbb R$ (or e.g., infinite-dimensional space over $Bbb Q$) in the apparent way.
However, $(1,0,0)$ and $(i,0,0)$ are $Bbb R$-linearly independant and $Bbb C$-linearly dependant.
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
$endgroup$
add a comment |
$begingroup$
Your assumption that $ix_1+0x_2+0x_3=0$ is wrong.
Your three vectors are linearly independent, no matter if we view $Bbb C^3$ as three-dimensional space over $Bbb C$, or six-dimensional space over $Bbb R$ (or e.g., infinite-dimensional space over $Bbb Q$) in the apparent way.
However, $(1,0,0)$ and $(i,0,0)$ are $Bbb R$-linearly independant and $Bbb C$-linearly dependant.
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
$endgroup$
add a comment |
$begingroup$
Your assumption that $ix_1+0x_2+0x_3=0$ is wrong.
Your three vectors are linearly independent, no matter if we view $Bbb C^3$ as three-dimensional space over $Bbb C$, or six-dimensional space over $Bbb R$ (or e.g., infinite-dimensional space over $Bbb Q$) in the apparent way.
However, $(1,0,0)$ and $(i,0,0)$ are $Bbb R$-linearly independant and $Bbb C$-linearly dependant.
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
$endgroup$
Your assumption that $ix_1+0x_2+0x_3=0$ is wrong.
Your three vectors are linearly independent, no matter if we view $Bbb C^3$ as three-dimensional space over $Bbb C$, or six-dimensional space over $Bbb R$ (or e.g., infinite-dimensional space over $Bbb Q$) in the apparent way.
However, $(1,0,0)$ and $(i,0,0)$ are $Bbb R$-linearly independant and $Bbb C$-linearly dependant.
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
answered Jan 12 at 17:45
Hagen von EitzenHagen von Eitzen
278k22269498
278k22269498
add a comment |
add a comment |
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$begingroup$
"...for the scalars..." . What scalars??
$endgroup$
– DonAntonio
Jan 12 at 17:44
5
$begingroup$
The given set of scalars doesn't satisfy (1).
$endgroup$
– Thomas Shelby
Jan 12 at 17:45
1
$begingroup$
Hint: $sumalpha_jx_j=(alpha_1,alpha_2,alpha_3)$.
$endgroup$
– David C. Ullrich
Jan 12 at 18:03