Conjugation map on a complex vector space Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)find all the conjugations on $C$ and $C^2$Extending a real vector space into a complex vector space using a linear mapJordan block in a matrix with a complex eigenvalueJordan-Chevalley decomposition of $T$ acting on $k[T]/(pi(T)^e)$Show that $S$ is a real vector space using the standard operations on $mathbbR^3$ (what are “standard operations” on $mathbbR^3$?)Range Space and Null Space proof, show that $y ∈ R(P)$ if and only if $y^T x = 0$ for every $x ∈ N (P)$Cayley-Dickson construction: a general rule for multiplying imaginary units?Is a complex vector space closed under complex conjugation?Restricted conjugation map, proof should follow from the conjugation map?Inner product over complex vector space and conjugationReal n-by-n Matrices…
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Conjugation map on a complex vector space
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)find all the conjugations on $C$ and $C^2$Extending a real vector space into a complex vector space using a linear mapJordan block in a matrix with a complex eigenvalueJordan-Chevalley decomposition of $T$ acting on $k[T]/(pi(T)^e)$Show that $S$ is a real vector space using the standard operations on $mathbbR^3$ (what are “standard operations” on $mathbbR^3$?)Range Space and Null Space proof, show that $y ∈ R(P)$ if and only if $y^T x = 0$ for every $x ∈ N (P)$Cayley-Dickson construction: a general rule for multiplying imaginary units?Is a complex vector space closed under complex conjugation?Restricted conjugation map, proof should follow from the conjugation map?Inner product over complex vector space and conjugationReal n-by-n Matrices…
$begingroup$
Let $V$ be a complex vector space and $J:Vto V$ a map with the following properties:
(i) $J(v+w)=J(v)+J(w)$
(ii) $J(cv)=overlinecJ(v)$
(iii) $J(J(v))=v$
for all $v,win V$ and $cinmathbbC$. Put $W=, J(v)=v$; this is a real vector space with respect to the operations in $V$ (I have shown this). From here, I am asked to show:
$bullet$ For each $vin V$, there exist unique vectors $w,uin W$ such that $v=w+iu$.
The uniqueness is very straight-forward once you have the decomposition. I've been trying the usual kinds of tricks, like writing $v=bigl(v-J(v)bigr)+J(v)$, but this clearly doesn't work. And I understand morally what's happening: we're decomposing $V$ into its formal real part and formal imaginary part. I just don't see how to get the decomposition.
Thanks
linear-algebra abstract-algebra
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
$endgroup$
add a comment |
$begingroup$
Let $V$ be a complex vector space and $J:Vto V$ a map with the following properties:
(i) $J(v+w)=J(v)+J(w)$
(ii) $J(cv)=overlinecJ(v)$
(iii) $J(J(v))=v$
for all $v,win V$ and $cinmathbbC$. Put $W=, J(v)=v$; this is a real vector space with respect to the operations in $V$ (I have shown this). From here, I am asked to show:
$bullet$ For each $vin V$, there exist unique vectors $w,uin W$ such that $v=w+iu$.
The uniqueness is very straight-forward once you have the decomposition. I've been trying the usual kinds of tricks, like writing $v=bigl(v-J(v)bigr)+J(v)$, but this clearly doesn't work. And I understand morally what's happening: we're decomposing $V$ into its formal real part and formal imaginary part. I just don't see how to get the decomposition.
Thanks
linear-algebra abstract-algebra
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
$endgroup$
$begingroup$
You need to show $W=mathbbR$ because then you know that $mathbbC=a+ib,, $
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 2:39
$begingroup$
I don't think that $W=mathbbR$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space.
$endgroup$
– Bey
Dec 11 '14 at 4:31
$begingroup$
I'm so sorry. I mistook $V$ for $mathbbC$
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 5:43
$begingroup$
Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation.
$endgroup$
– Michael Albanese
Apr 3 '15 at 2:54
$begingroup$
Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text.
$endgroup$
– Bey
Apr 3 '15 at 11:36
add a comment |
$begingroup$
Let $V$ be a complex vector space and $J:Vto V$ a map with the following properties:
(i) $J(v+w)=J(v)+J(w)$
(ii) $J(cv)=overlinecJ(v)$
(iii) $J(J(v))=v$
for all $v,win V$ and $cinmathbbC$. Put $W=, J(v)=v$; this is a real vector space with respect to the operations in $V$ (I have shown this). From here, I am asked to show:
$bullet$ For each $vin V$, there exist unique vectors $w,uin W$ such that $v=w+iu$.
The uniqueness is very straight-forward once you have the decomposition. I've been trying the usual kinds of tricks, like writing $v=bigl(v-J(v)bigr)+J(v)$, but this clearly doesn't work. And I understand morally what's happening: we're decomposing $V$ into its formal real part and formal imaginary part. I just don't see how to get the decomposition.
Thanks
linear-algebra abstract-algebra
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
$endgroup$
Let $V$ be a complex vector space and $J:Vto V$ a map with the following properties:
(i) $J(v+w)=J(v)+J(w)$
(ii) $J(cv)=overlinecJ(v)$
(iii) $J(J(v))=v$
for all $v,win V$ and $cinmathbbC$. Put $W=, J(v)=v$; this is a real vector space with respect to the operations in $V$ (I have shown this). From here, I am asked to show:
$bullet$ For each $vin V$, there exist unique vectors $w,uin W$ such that $v=w+iu$.
The uniqueness is very straight-forward once you have the decomposition. I've been trying the usual kinds of tricks, like writing $v=bigl(v-J(v)bigr)+J(v)$, but this clearly doesn't work. And I understand morally what's happening: we're decomposing $V$ into its formal real part and formal imaginary part. I just don't see how to get the decomposition.
Thanks
linear-algebra abstract-algebra
linear-algebra abstract-algebra
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
edited Dec 11 '14 at 19:24
Michael Albanese
64.9k1599316
64.9k1599316
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
asked Dec 11 '14 at 2:34
BeyBey
1,9801638
1,9801638
$begingroup$
You need to show $W=mathbbR$ because then you know that $mathbbC=a+ib,, $
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 2:39
$begingroup$
I don't think that $W=mathbbR$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space.
$endgroup$
– Bey
Dec 11 '14 at 4:31
$begingroup$
I'm so sorry. I mistook $V$ for $mathbbC$
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 5:43
$begingroup$
Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation.
$endgroup$
– Michael Albanese
Apr 3 '15 at 2:54
$begingroup$
Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text.
$endgroup$
– Bey
Apr 3 '15 at 11:36
add a comment |
$begingroup$
You need to show $W=mathbbR$ because then you know that $mathbbC=a+ib,, $
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 2:39
$begingroup$
I don't think that $W=mathbbR$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space.
$endgroup$
– Bey
Dec 11 '14 at 4:31
$begingroup$
I'm so sorry. I mistook $V$ for $mathbbC$
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 5:43
$begingroup$
Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation.
$endgroup$
– Michael Albanese
Apr 3 '15 at 2:54
$begingroup$
Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text.
$endgroup$
– Bey
Apr 3 '15 at 11:36
$begingroup$
You need to show $W=mathbbR$ because then you know that $mathbbC=a+ib,, $
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 2:39
$begingroup$
You need to show $W=mathbbR$ because then you know that $mathbbC=a+ib,, $
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 2:39
$begingroup$
I don't think that $W=mathbbR$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space.
$endgroup$
– Bey
Dec 11 '14 at 4:31
$begingroup$
I don't think that $W=mathbbR$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space.
$endgroup$
– Bey
Dec 11 '14 at 4:31
$begingroup$
I'm so sorry. I mistook $V$ for $mathbbC$
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 5:43
$begingroup$
I'm so sorry. I mistook $V$ for $mathbbC$
$endgroup$
– Swapnil Tripathi
Dec 11 '14 at 5:43
$begingroup$
Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation.
$endgroup$
– Michael Albanese
Apr 3 '15 at 2:54
$begingroup$
Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation.
$endgroup$
– Michael Albanese
Apr 3 '15 at 2:54
$begingroup$
Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text.
$endgroup$
– Bey
Apr 3 '15 at 11:36
$begingroup$
Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text.
$endgroup$
– Bey
Apr 3 '15 at 11:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given $z in mathbbC$, we have $operatornameRe(z) = frac12(z + barz)$ and $operatornameIm(z) = frac12i(z - barz)$. Let's try the analogous thing, where in place of $mathbbC$ we have $V$, and instead of conjugation, we have $J$.
Let $w = frac12(v + J(v))$ and $u = frac12i(v - J(v))$; note, we're allowed to multiply by $frac12i$ in the expression for $u$ because $V$ is a complex vector space. Now note that
$$J(w) = Jleft(frac12(v + J(v))right) = frac12(J(v)+J(J(v))) = frac12(J(v) + v) = w$$
and
$$J(u) = Jleft(frac12i(v - J(v))right) = frac-12i(J(v)-J(J(v))) =frac-12i(J(v) - v) = u.$$
So $w, u in W$ and
$$w + iu = frac12(v + J(v)) + ileft(frac12i(v - J(v))right) = frac12(v + J(v)) + frac12(v - J(v)) = v.$$
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
$endgroup$
add a comment |
$begingroup$
For completeness, in order to get uniqueness, I would make an argument as follows.
If $z = x + iy = x' + iy'$ where $x, x', y, y' in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' in W$, so $i(y' - y) in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) in W$, $y' - y = 0$, which implies $x = x', y = y'$.
edited Mar 27 at 18:21
Maurice P
1,4561732
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Given $z in mathbbC$, we have $operatornameRe(z) = frac12(z + barz)$ and $operatornameIm(z) = frac12i(z - barz)$. Let's try the analogous thing, where in place of $mathbbC$ we have $V$, and instead of conjugation, we have $J$.
Let $w = frac12(v + J(v))$ and $u = frac12i(v - J(v))$; note, we're allowed to multiply by $frac12i$ in the expression for $u$ because $V$ is a complex vector space. Now note that
$$J(w) = Jleft(frac12(v + J(v))right) = frac12(J(v)+J(J(v))) = frac12(J(v) + v) = w$$
and
$$J(u) = Jleft(frac12i(v - J(v))right) = frac-12i(J(v)-J(J(v))) =frac-12i(J(v) - v) = u.$$
So $w, u in W$ and
$$w + iu = frac12(v + J(v)) + ileft(frac12i(v - J(v))right) = frac12(v + J(v)) + frac12(v - J(v)) = v.$$
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
$endgroup$
add a comment |
$begingroup$
Given $z in mathbbC$, we have $operatornameRe(z) = frac12(z + barz)$ and $operatornameIm(z) = frac12i(z - barz)$. Let's try the analogous thing, where in place of $mathbbC$ we have $V$, and instead of conjugation, we have $J$.
Let $w = frac12(v + J(v))$ and $u = frac12i(v - J(v))$; note, we're allowed to multiply by $frac12i$ in the expression for $u$ because $V$ is a complex vector space. Now note that
$$J(w) = Jleft(frac12(v + J(v))right) = frac12(J(v)+J(J(v))) = frac12(J(v) + v) = w$$
and
$$J(u) = Jleft(frac12i(v - J(v))right) = frac-12i(J(v)-J(J(v))) =frac-12i(J(v) - v) = u.$$
So $w, u in W$ and
$$w + iu = frac12(v + J(v)) + ileft(frac12i(v - J(v))right) = frac12(v + J(v)) + frac12(v - J(v)) = v.$$
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
$endgroup$
add a comment |
$begingroup$
Given $z in mathbbC$, we have $operatornameRe(z) = frac12(z + barz)$ and $operatornameIm(z) = frac12i(z - barz)$. Let's try the analogous thing, where in place of $mathbbC$ we have $V$, and instead of conjugation, we have $J$.
Let $w = frac12(v + J(v))$ and $u = frac12i(v - J(v))$; note, we're allowed to multiply by $frac12i$ in the expression for $u$ because $V$ is a complex vector space. Now note that
$$J(w) = Jleft(frac12(v + J(v))right) = frac12(J(v)+J(J(v))) = frac12(J(v) + v) = w$$
and
$$J(u) = Jleft(frac12i(v - J(v))right) = frac-12i(J(v)-J(J(v))) =frac-12i(J(v) - v) = u.$$
So $w, u in W$ and
$$w + iu = frac12(v + J(v)) + ileft(frac12i(v - J(v))right) = frac12(v + J(v)) + frac12(v - J(v)) = v.$$
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
$endgroup$
Given $z in mathbbC$, we have $operatornameRe(z) = frac12(z + barz)$ and $operatornameIm(z) = frac12i(z - barz)$. Let's try the analogous thing, where in place of $mathbbC$ we have $V$, and instead of conjugation, we have $J$.
Let $w = frac12(v + J(v))$ and $u = frac12i(v - J(v))$; note, we're allowed to multiply by $frac12i$ in the expression for $u$ because $V$ is a complex vector space. Now note that
$$J(w) = Jleft(frac12(v + J(v))right) = frac12(J(v)+J(J(v))) = frac12(J(v) + v) = w$$
and
$$J(u) = Jleft(frac12i(v - J(v))right) = frac-12i(J(v)-J(J(v))) =frac-12i(J(v) - v) = u.$$
So $w, u in W$ and
$$w + iu = frac12(v + J(v)) + ileft(frac12i(v - J(v))right) = frac12(v + J(v)) + frac12(v - J(v)) = v.$$
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
answered Dec 11 '14 at 19:23
Michael AlbaneseMichael Albanese
64.9k1599316
64.9k1599316
add a comment |
add a comment |
$begingroup$
For completeness, in order to get uniqueness, I would make an argument as follows.
If $z = x + iy = x' + iy'$ where $x, x', y, y' in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' in W$, so $i(y' - y) in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) in W$, $y' - y = 0$, which implies $x = x', y = y'$.
edited Mar 27 at 18:21
Maurice P
1,4561732
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
$endgroup$
add a comment |
$begingroup$
For completeness, in order to get uniqueness, I would make an argument as follows.
If $z = x + iy = x' + iy'$ where $x, x', y, y' in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' in W$, so $i(y' - y) in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) in W$, $y' - y = 0$, which implies $x = x', y = y'$.
edited Mar 27 at 18:21
Maurice P
1,4561732
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
$endgroup$
add a comment |
$begingroup$
For completeness, in order to get uniqueness, I would make an argument as follows.
If $z = x + iy = x' + iy'$ where $x, x', y, y' in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' in W$, so $i(y' - y) in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) in W$, $y' - y = 0$, which implies $x = x', y = y'$.
edited Mar 27 at 18:21
Maurice P
1,4561732
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
$endgroup$
For completeness, in order to get uniqueness, I would make an argument as follows.
If $z = x + iy = x' + iy'$ where $x, x', y, y' in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' in W$, so $i(y' - y) in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) in W$, $y' - y = 0$, which implies $x = x', y = y'$.
edited Mar 27 at 18:21
Maurice P
1,4561732
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
edited Mar 27 at 18:21
Maurice P
1,4561732
edited Mar 27 at 18:21
Maurice P
1,4561732
edited Mar 27 at 18:21
Maurice P
1,4561732
1,4561732
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
answered May 20 '16 at 23:34
David Warren KatzDavid Warren Katz
598315
598315
add a comment |
add a comment |
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You need to show $W=mathbbR$ because then you know that $mathbbC=a+ib,, $
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– Swapnil Tripathi
Dec 11 '14 at 2:39
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I don't think that $W=mathbbR$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space.
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– Bey
Dec 11 '14 at 4:31
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I'm so sorry. I mistook $V$ for $mathbbC$
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– Swapnil Tripathi
Dec 11 '14 at 5:43
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Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation.
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– Michael Albanese
Apr 3 '15 at 2:54
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Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text.
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– Bey
Apr 3 '15 at 11:36