Mixing
Is there a set of three linearly independent vectors in $mathbb{C}^2$?
$begingroup$
I claim the set of three linearly independent vectors in $mathbb{C}^2$ to be empty.
But I'm not sure if the following proof is correct.
Let $x$ and $y$ be linearly independent vectors in $mathbb{C}^2$ over the field $mathbb{C}$.
Assume that $z neq 0$ in $mathbb{C}^2$ is not a linear combination of $x$ and $y$.
Then, for all $alpha_1, alpha_2$ in $mathbb{C}$, we have that
$$
begin{eqnarray}
z & = & z_1 (1, 0) + z_2 (0, 1) \
& neq & alpha_1 x + alpha_2 y \
& = & (alpha_1 x_1 + alpha_2 y_1)(1, 0) + (alpha_1 x_2 + alpha_2 x_2)(0, 1).
end{eqnarray}
$$
This requires that either $z_1 neq (alpha_1x_1 + alpha_2y_1)$, or $z_2 neq (alpha_1x_2 + alpha_2y_2)$, or both.
But if $z_1 neq (alpha_1x_1 + alpha_2y_1)$, for all $alpha_1, alpha_2$, this can only be the case if $x_1 = y_1 = 0$. This, in turn, causes ${x, y}$ to be linearly dependent, contradicting the assumption. A similar argument applies for $z_2$. Hence, either $z = 0$ or $z$ is a linear combination of $x$ and $y$.
linear-algebra proof-verification
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
$endgroup$
|
show 1 more comment
$begingroup$
I claim the set of three linearly independent vectors in $mathbb{C}^2$ to be empty.
But I'm not sure if the following proof is correct.
Let $x$ and $y$ be linearly independent vectors in $mathbb{C}^2$ over the field $mathbb{C}$.
Assume that $z neq 0$ in $mathbb{C}^2$ is not a linear combination of $x$ and $y$.
Then, for all $alpha_1, alpha_2$ in $mathbb{C}$, we have that
$$
begin{eqnarray}
z & = & z_1 (1, 0) + z_2 (0, 1) \
& neq & alpha_1 x + alpha_2 y \
& = & (alpha_1 x_1 + alpha_2 y_1)(1, 0) + (alpha_1 x_2 + alpha_2 x_2)(0, 1).
end{eqnarray}
$$
This requires that either $z_1 neq (alpha_1x_1 + alpha_2y_1)$, or $z_2 neq (alpha_1x_2 + alpha_2y_2)$, or both.
But if $z_1 neq (alpha_1x_1 + alpha_2y_1)$, for all $alpha_1, alpha_2$, this can only be the case if $x_1 = y_1 = 0$. This, in turn, causes ${x, y}$ to be linearly dependent, contradicting the assumption. A similar argument applies for $z_2$. Hence, either $z = 0$ or $z$ is a linear combination of $x$ and $y$.
linear-algebra proof-verification
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
$endgroup$
2
$begingroup$
Your notation is confusing. You start with $x,y$ linearly independent, but then mix it with $(1,0)$ and $(0,1)$?
$endgroup$
– Dietrich Burde
Jan 16 at 21:20
$begingroup$
You didn't really prove the statement of last paragraph after "But if".
$endgroup$
– coffeemath
Jan 16 at 21:21
4
$begingroup$
$mathbb{C}^2$ has $text{dim}$ $4$ as a vector space over $mathbb{R},$ but $text{dim}$ $2$ as a vector space over $mathbb{C}.$
$endgroup$
– Bumblebee
Jan 16 at 21:21
1
$begingroup$
What is your scalar field? $mathbb C$ or $mathbb R$? You should be explicit and say, since it affects the answer.
$endgroup$
– MPW
Jan 16 at 21:30
1
$begingroup$
@MaxHerrmann The phrase" Then, for all $alpha_1,alpha_2$" is not true. Take $alpha_1=alpha_2=0$ and $z=0$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:32
|
show 1 more comment
$begingroup$
I claim the set of three linearly independent vectors in $mathbb{C}^2$ to be empty.
But I'm not sure if the following proof is correct.
Let $x$ and $y$ be linearly independent vectors in $mathbb{C}^2$ over the field $mathbb{C}$.
Assume that $z neq 0$ in $mathbb{C}^2$ is not a linear combination of $x$ and $y$.
Then, for all $alpha_1, alpha_2$ in $mathbb{C}$, we have that
$$
begin{eqnarray}
z & = & z_1 (1, 0) + z_2 (0, 1) \
& neq & alpha_1 x + alpha_2 y \
& = & (alpha_1 x_1 + alpha_2 y_1)(1, 0) + (alpha_1 x_2 + alpha_2 x_2)(0, 1).
end{eqnarray}
$$
This requires that either $z_1 neq (alpha_1x_1 + alpha_2y_1)$, or $z_2 neq (alpha_1x_2 + alpha_2y_2)$, or both.
But if $z_1 neq (alpha_1x_1 + alpha_2y_1)$, for all $alpha_1, alpha_2$, this can only be the case if $x_1 = y_1 = 0$. This, in turn, causes ${x, y}$ to be linearly dependent, contradicting the assumption. A similar argument applies for $z_2$. Hence, either $z = 0$ or $z$ is a linear combination of $x$ and $y$.
linear-algebra proof-verification
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
$endgroup$
I claim the set of three linearly independent vectors in $mathbb{C}^2$ to be empty.
But I'm not sure if the following proof is correct.
Let $x$ and $y$ be linearly independent vectors in $mathbb{C}^2$ over the field $mathbb{C}$.
Assume that $z neq 0$ in $mathbb{C}^2$ is not a linear combination of $x$ and $y$.
Then, for all $alpha_1, alpha_2$ in $mathbb{C}$, we have that
$$
begin{eqnarray}
z & = & z_1 (1, 0) + z_2 (0, 1) \
& neq & alpha_1 x + alpha_2 y \
& = & (alpha_1 x_1 + alpha_2 y_1)(1, 0) + (alpha_1 x_2 + alpha_2 x_2)(0, 1).
end{eqnarray}
$$
This requires that either $z_1 neq (alpha_1x_1 + alpha_2y_1)$, or $z_2 neq (alpha_1x_2 + alpha_2y_2)$, or both.
But if $z_1 neq (alpha_1x_1 + alpha_2y_1)$, for all $alpha_1, alpha_2$, this can only be the case if $x_1 = y_1 = 0$. This, in turn, causes ${x, y}$ to be linearly dependent, contradicting the assumption. A similar argument applies for $z_2$. Hence, either $z = 0$ or $z$ is a linear combination of $x$ and $y$.
linear-algebra proof-verification
linear-algebra proof-verification
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
edited Jan 17 at 8:10
Max Herrmann
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
asked Jan 16 at 21:17
Max HerrmannMax Herrmann
704418
704418
2
$begingroup$
Your notation is confusing. You start with $x,y$ linearly independent, but then mix it with $(1,0)$ and $(0,1)$?
$endgroup$
– Dietrich Burde
Jan 16 at 21:20
$begingroup$
You didn't really prove the statement of last paragraph after "But if".
$endgroup$
– coffeemath
Jan 16 at 21:21
4
$begingroup$
$mathbb{C}^2$ has $text{dim}$ $4$ as a vector space over $mathbb{R},$ but $text{dim}$ $2$ as a vector space over $mathbb{C}.$
$endgroup$
– Bumblebee
Jan 16 at 21:21
1
$begingroup$
What is your scalar field? $mathbb C$ or $mathbb R$? You should be explicit and say, since it affects the answer.
$endgroup$
– MPW
Jan 16 at 21:30
1
$begingroup$
@MaxHerrmann The phrase" Then, for all $alpha_1,alpha_2$" is not true. Take $alpha_1=alpha_2=0$ and $z=0$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:32
|
show 1 more comment
2
$begingroup$
Your notation is confusing. You start with $x,y$ linearly independent, but then mix it with $(1,0)$ and $(0,1)$?
$endgroup$
– Dietrich Burde
Jan 16 at 21:20
$begingroup$
You didn't really prove the statement of last paragraph after "But if".
$endgroup$
– coffeemath
Jan 16 at 21:21
4
$begingroup$
$mathbb{C}^2$ has $text{dim}$ $4$ as a vector space over $mathbb{R},$ but $text{dim}$ $2$ as a vector space over $mathbb{C}.$
$endgroup$
– Bumblebee
Jan 16 at 21:21
1
$begingroup$
What is your scalar field? $mathbb C$ or $mathbb R$? You should be explicit and say, since it affects the answer.
$endgroup$
– MPW
Jan 16 at 21:30
1
$begingroup$
@MaxHerrmann The phrase" Then, for all $alpha_1,alpha_2$" is not true. Take $alpha_1=alpha_2=0$ and $z=0$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:32
2
2
$begingroup$
Your notation is confusing. You start with $x,y$ linearly independent, but then mix it with $(1,0)$ and $(0,1)$?
$endgroup$
– Dietrich Burde
Jan 16 at 21:20
$begingroup$
Your notation is confusing. You start with $x,y$ linearly independent, but then mix it with $(1,0)$ and $(0,1)$?
$endgroup$
– Dietrich Burde
Jan 16 at 21:20
$begingroup$
You didn't really prove the statement of last paragraph after "But if".
$endgroup$
– coffeemath
Jan 16 at 21:21
$begingroup$
You didn't really prove the statement of last paragraph after "But if".
$endgroup$
– coffeemath
Jan 16 at 21:21
4
4
$begingroup$
$mathbb{C}^2$ has $text{dim}$ $4$ as a vector space over $mathbb{R},$ but $text{dim}$ $2$ as a vector space over $mathbb{C}.$
$endgroup$
– Bumblebee
Jan 16 at 21:21
$begingroup$
$mathbb{C}^2$ has $text{dim}$ $4$ as a vector space over $mathbb{R},$ but $text{dim}$ $2$ as a vector space over $mathbb{C}.$
$endgroup$
– Bumblebee
Jan 16 at 21:21
1
1
$begingroup$
What is your scalar field? $mathbb C$ or $mathbb R$? You should be explicit and say, since it affects the answer.
$endgroup$
– MPW
Jan 16 at 21:30
$begingroup$
What is your scalar field? $mathbb C$ or $mathbb R$? You should be explicit and say, since it affects the answer.
$endgroup$
– MPW
Jan 16 at 21:30
1
1
$begingroup$
@MaxHerrmann The phrase" Then, for all $alpha_1,alpha_2$" is not true. Take $alpha_1=alpha_2=0$ and $z=0$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:32
$begingroup$
@MaxHerrmann The phrase" Then, for all $alpha_1,alpha_2$" is not true. Take $alpha_1=alpha_2=0$ and $z=0$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:32
|
show 1 more comment
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$begingroup$
Your notation is confusing. You start with $x,y$ linearly independent, but then mix it with $(1,0)$ and $(0,1)$?
$endgroup$
– Dietrich Burde
Jan 16 at 21:20
$begingroup$
You didn't really prove the statement of last paragraph after "But if".
$endgroup$
– coffeemath
Jan 16 at 21:21
4
$begingroup$
$mathbb{C}^2$ has $text{dim}$ $4$ as a vector space over $mathbb{R},$ but $text{dim}$ $2$ as a vector space over $mathbb{C}.$
$endgroup$
– Bumblebee
Jan 16 at 21:21
1
$begingroup$
What is your scalar field? $mathbb C$ or $mathbb R$? You should be explicit and say, since it affects the answer.
$endgroup$
– MPW
Jan 16 at 21:30
1
$begingroup$
@MaxHerrmann The phrase" Then, for all $alpha_1,alpha_2$" is not true. Take $alpha_1=alpha_2=0$ and $z=0$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:32