Halmos Finite-Dimensional Vector Spaces: Ordered pairs of complex numbers
$begingroup$
Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 1, exercise 7.c:
What happens (in both the preceding cases) if we consider ordered
pairs of complex numbers instead?
Question asked is if
- the set of ordered pairs $(a, b) in mathbb{C}^2$
- addition $(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)$
- multiplication $(a_1, b_1) cdot (a_2, b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1)$
form a field.
I decided to check if the above given object has all the properties a field needs.
Currently, I'm stuck at the question if a multiplicative inverse exists for every $(a, b)$. This would require that
$$
(a, b)^{-1}(a,b) = (1, 0),
$$
since I found $(1, 0)$ to be the neutral element of multiplication.
I ended up with the following set of linear equations:
$$
left(
begin{array}{cccc}
alpha_1 & -alpha_2 & -beta_1 & beta_2 \
alpha_2 & alpha_1 & -beta_2 & -beta_1 \
beta_1 & -beta_2 & alpha_1 & -alpha_2 \
beta_1 & beta_2 & alpha_1 & alpha_2
end{array}
right)
left(
begin{array}{c}
gamma_1 \
gamma_2 \
delta_1 \
delta_2
end{array}
right)
=
left(
begin{array}{c}
1 \
0 \
0 \
0
end{array}
right),
$$
where $(a, b) = (alpha_1 + i beta_1, alpha_2 + i beta_2)$, and $(a, b)^{-1} = (gamma_1 + i delta_1, gamma_2 + i delta_2)$.
Applying LAPLACE's formula, I found the expression for the determinant of the matrix to be
$$
2(alpha_1^2 + alpha_2^2 + beta_1^2 + beta_2^2)(alpha_1alpha_2 + beta_1beta_2).
$$
Then, if $alpha_1alpha_2 + beta_1beta_2 = 0$, the determinant is $0$ and the solution to the system of equations is not unique.
May I conclude that there are cases when no (unique) inverse exists, and the above given object is not a field?
abstract-algebra
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
$endgroup$
|
show 2 more comments
$begingroup$
Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 1, exercise 7.c:
What happens (in both the preceding cases) if we consider ordered
pairs of complex numbers instead?
Question asked is if
- the set of ordered pairs $(a, b) in mathbb{C}^2$
- addition $(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)$
- multiplication $(a_1, b_1) cdot (a_2, b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1)$
form a field.
I decided to check if the above given object has all the properties a field needs.
Currently, I'm stuck at the question if a multiplicative inverse exists for every $(a, b)$. This would require that
$$
(a, b)^{-1}(a,b) = (1, 0),
$$
since I found $(1, 0)$ to be the neutral element of multiplication.
I ended up with the following set of linear equations:
$$
left(
begin{array}{cccc}
alpha_1 & -alpha_2 & -beta_1 & beta_2 \
alpha_2 & alpha_1 & -beta_2 & -beta_1 \
beta_1 & -beta_2 & alpha_1 & -alpha_2 \
beta_1 & beta_2 & alpha_1 & alpha_2
end{array}
right)
left(
begin{array}{c}
gamma_1 \
gamma_2 \
delta_1 \
delta_2
end{array}
right)
=
left(
begin{array}{c}
1 \
0 \
0 \
0
end{array}
right),
$$
where $(a, b) = (alpha_1 + i beta_1, alpha_2 + i beta_2)$, and $(a, b)^{-1} = (gamma_1 + i delta_1, gamma_2 + i delta_2)$.
Applying LAPLACE's formula, I found the expression for the determinant of the matrix to be
$$
2(alpha_1^2 + alpha_2^2 + beta_1^2 + beta_2^2)(alpha_1alpha_2 + beta_1beta_2).
$$
Then, if $alpha_1alpha_2 + beta_1beta_2 = 0$, the determinant is $0$ and the solution to the system of equations is not unique.
May I conclude that there are cases when no (unique) inverse exists, and the above given object is not a field?
abstract-algebra
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
$endgroup$
6
$begingroup$
Try $(1,i)cdot(1,-i)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 16:55
$begingroup$
Thank you @LordSharktheUnknown! This states that there are complex pairs s.t. $a cdot b = 0$. But what does that imply?
$endgroup$
– Max Herrmann
Jan 1 at 17:10
1
$begingroup$
In a field, if $a,bneq0$, then $abneq0$.
$endgroup$
– José Carlos Santos
Jan 1 at 17:13
2
$begingroup$
An easier way than the one suggested by Lord Shark the Unknown? I don't think so.
$endgroup$
– José Carlos Santos
Jan 1 at 17:20
2
$begingroup$
Last time I checked there were infinitely many complex numbers, so the tag finite fields was inappropriate.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 5:51
|
show 2 more comments
$begingroup$
Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 1, exercise 7.c:
What happens (in both the preceding cases) if we consider ordered
pairs of complex numbers instead?
Question asked is if
- the set of ordered pairs $(a, b) in mathbb{C}^2$
- addition $(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)$
- multiplication $(a_1, b_1) cdot (a_2, b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1)$
form a field.
I decided to check if the above given object has all the properties a field needs.
Currently, I'm stuck at the question if a multiplicative inverse exists for every $(a, b)$. This would require that
$$
(a, b)^{-1}(a,b) = (1, 0),
$$
since I found $(1, 0)$ to be the neutral element of multiplication.
I ended up with the following set of linear equations:
$$
left(
begin{array}{cccc}
alpha_1 & -alpha_2 & -beta_1 & beta_2 \
alpha_2 & alpha_1 & -beta_2 & -beta_1 \
beta_1 & -beta_2 & alpha_1 & -alpha_2 \
beta_1 & beta_2 & alpha_1 & alpha_2
end{array}
right)
left(
begin{array}{c}
gamma_1 \
gamma_2 \
delta_1 \
delta_2
end{array}
right)
=
left(
begin{array}{c}
1 \
0 \
0 \
0
end{array}
right),
$$
where $(a, b) = (alpha_1 + i beta_1, alpha_2 + i beta_2)$, and $(a, b)^{-1} = (gamma_1 + i delta_1, gamma_2 + i delta_2)$.
Applying LAPLACE's formula, I found the expression for the determinant of the matrix to be
$$
2(alpha_1^2 + alpha_2^2 + beta_1^2 + beta_2^2)(alpha_1alpha_2 + beta_1beta_2).
$$
Then, if $alpha_1alpha_2 + beta_1beta_2 = 0$, the determinant is $0$ and the solution to the system of equations is not unique.
May I conclude that there are cases when no (unique) inverse exists, and the above given object is not a field?
abstract-algebra
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
$endgroup$
Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 1, exercise 7.c:
What happens (in both the preceding cases) if we consider ordered
pairs of complex numbers instead?
Question asked is if
- the set of ordered pairs $(a, b) in mathbb{C}^2$
- addition $(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)$
- multiplication $(a_1, b_1) cdot (a_2, b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1)$
form a field.
I decided to check if the above given object has all the properties a field needs.
Currently, I'm stuck at the question if a multiplicative inverse exists for every $(a, b)$. This would require that
$$
(a, b)^{-1}(a,b) = (1, 0),
$$
since I found $(1, 0)$ to be the neutral element of multiplication.
I ended up with the following set of linear equations:
$$
left(
begin{array}{cccc}
alpha_1 & -alpha_2 & -beta_1 & beta_2 \
alpha_2 & alpha_1 & -beta_2 & -beta_1 \
beta_1 & -beta_2 & alpha_1 & -alpha_2 \
beta_1 & beta_2 & alpha_1 & alpha_2
end{array}
right)
left(
begin{array}{c}
gamma_1 \
gamma_2 \
delta_1 \
delta_2
end{array}
right)
=
left(
begin{array}{c}
1 \
0 \
0 \
0
end{array}
right),
$$
where $(a, b) = (alpha_1 + i beta_1, alpha_2 + i beta_2)$, and $(a, b)^{-1} = (gamma_1 + i delta_1, gamma_2 + i delta_2)$.
Applying LAPLACE's formula, I found the expression for the determinant of the matrix to be
$$
2(alpha_1^2 + alpha_2^2 + beta_1^2 + beta_2^2)(alpha_1alpha_2 + beta_1beta_2).
$$
Then, if $alpha_1alpha_2 + beta_1beta_2 = 0$, the determinant is $0$ and the solution to the system of equations is not unique.
May I conclude that there are cases when no (unique) inverse exists, and the above given object is not a field?
abstract-algebra
abstract-algebra
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
edited Jan 2 at 5:50
Jyrki Lahtonen
110k13171378
110k13171378
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
asked Jan 1 at 16:52
Max HerrmannMax Herrmann
704418
704418
6
$begingroup$
Try $(1,i)cdot(1,-i)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 16:55
$begingroup$
Thank you @LordSharktheUnknown! This states that there are complex pairs s.t. $a cdot b = 0$. But what does that imply?
$endgroup$
– Max Herrmann
Jan 1 at 17:10
1
$begingroup$
In a field, if $a,bneq0$, then $abneq0$.
$endgroup$
– José Carlos Santos
Jan 1 at 17:13
2
$begingroup$
An easier way than the one suggested by Lord Shark the Unknown? I don't think so.
$endgroup$
– José Carlos Santos
Jan 1 at 17:20
2
$begingroup$
Last time I checked there were infinitely many complex numbers, so the tag finite fields was inappropriate.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 5:51
|
show 2 more comments
6
$begingroup$
Try $(1,i)cdot(1,-i)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 16:55
$begingroup$
Thank you @LordSharktheUnknown! This states that there are complex pairs s.t. $a cdot b = 0$. But what does that imply?
$endgroup$
– Max Herrmann
Jan 1 at 17:10
1
$begingroup$
In a field, if $a,bneq0$, then $abneq0$.
$endgroup$
– José Carlos Santos
Jan 1 at 17:13
2
$begingroup$
An easier way than the one suggested by Lord Shark the Unknown? I don't think so.
$endgroup$
– José Carlos Santos
Jan 1 at 17:20
2
$begingroup$
Last time I checked there were infinitely many complex numbers, so the tag finite fields was inappropriate.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 5:51
6
6
$begingroup$
Try $(1,i)cdot(1,-i)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 16:55
$begingroup$
Try $(1,i)cdot(1,-i)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 16:55
$begingroup$
Thank you @LordSharktheUnknown! This states that there are complex pairs s.t. $a cdot b = 0$. But what does that imply?
$endgroup$
– Max Herrmann
Jan 1 at 17:10
$begingroup$
Thank you @LordSharktheUnknown! This states that there are complex pairs s.t. $a cdot b = 0$. But what does that imply?
$endgroup$
– Max Herrmann
Jan 1 at 17:10
1
1
$begingroup$
In a field, if $a,bneq0$, then $abneq0$.
$endgroup$
– José Carlos Santos
Jan 1 at 17:13
$begingroup$
In a field, if $a,bneq0$, then $abneq0$.
$endgroup$
– José Carlos Santos
Jan 1 at 17:13
2
2
$begingroup$
An easier way than the one suggested by Lord Shark the Unknown? I don't think so.
$endgroup$
– José Carlos Santos
Jan 1 at 17:20
$begingroup$
An easier way than the one suggested by Lord Shark the Unknown? I don't think so.
$endgroup$
– José Carlos Santos
Jan 1 at 17:20
2
2
$begingroup$
Last time I checked there were infinitely many complex numbers, so the tag finite fields was inappropriate.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 5:51
$begingroup$
Last time I checked there were infinitely many complex numbers, so the tag finite fields was inappropriate.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 5:51
|
show 2 more comments
0
active
oldest
votes
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6
$begingroup$
Try $(1,i)cdot(1,-i)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 16:55
$begingroup$
Thank you @LordSharktheUnknown! This states that there are complex pairs s.t. $a cdot b = 0$. But what does that imply?
$endgroup$
– Max Herrmann
Jan 1 at 17:10
1
$begingroup$
In a field, if $a,bneq0$, then $abneq0$.
$endgroup$
– José Carlos Santos
Jan 1 at 17:13
2
$begingroup$
An easier way than the one suggested by Lord Shark the Unknown? I don't think so.
$endgroup$
– José Carlos Santos
Jan 1 at 17:20
2
$begingroup$
Last time I checked there were infinitely many complex numbers, so the tag finite fields was inappropriate.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 5:51