matrix homomorphism
$begingroup$
Prove that $ϕ$ is a homomorphism and describe its kernel.
$$
ϕ:ℝ→GL(2,ℝ),qquad
ϕ(x)=
begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
$$
where $x∈ℝ$
I have begun by saying
$$
y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
$$
$ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$
So I have calculated
$$
ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
$$
Is this right? And I don't know how to go from there. any help would be appreciated.
abstract-algebra group-homomorphism
edited Dec 3 '18 at 22:11
egreg
183k1486204
asked Dec 3 '18 at 21:56
jessicajessica
42
$endgroup$
add a comment |
$begingroup$
Prove that $ϕ$ is a homomorphism and describe its kernel.
$$
ϕ:ℝ→GL(2,ℝ),qquad
ϕ(x)=
begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
$$
where $x∈ℝ$
I have begun by saying
$$
y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
$$
$ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$
So I have calculated
$$
ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
$$
Is this right? And I don't know how to go from there. any help would be appreciated.
abstract-algebra group-homomorphism
edited Dec 3 '18 at 22:11
egreg
183k1486204
asked Dec 3 '18 at 21:56
jessicajessica
42
$endgroup$
add a comment |
$begingroup$
Prove that $ϕ$ is a homomorphism and describe its kernel.
$$
ϕ:ℝ→GL(2,ℝ),qquad
ϕ(x)=
begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
$$
where $x∈ℝ$
I have begun by saying
$$
y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
$$
$ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$
So I have calculated
$$
ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
$$
Is this right? And I don't know how to go from there. any help would be appreciated.
abstract-algebra group-homomorphism
edited Dec 3 '18 at 22:11
egreg
183k1486204
asked Dec 3 '18 at 21:56
jessicajessica
42
$endgroup$
Prove that $ϕ$ is a homomorphism and describe its kernel.
$$
ϕ:ℝ→GL(2,ℝ),qquad
ϕ(x)=
begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
$$
where $x∈ℝ$
I have begun by saying
$$
y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
$$
$ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$
So I have calculated
$$
ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
$$
Is this right? And I don't know how to go from there. any help would be appreciated.
abstract-algebra group-homomorphism
abstract-algebra group-homomorphism
edited Dec 3 '18 at 22:11
egreg
183k1486204
asked Dec 3 '18 at 21:56
jessicajessica
42
edited Dec 3 '18 at 22:11
egreg
183k1486204
asked Dec 3 '18 at 21:56
jessicajessica
42
edited Dec 3 '18 at 22:11
egreg
183k1486204
edited Dec 3 '18 at 22:11
egreg
183k1486204
edited Dec 3 '18 at 22:11
egreg
183k1486204
183k1486204
asked Dec 3 '18 at 21:56
jessicajessica
42
asked Dec 3 '18 at 21:56
jessicajessica
42
asked Dec 3 '18 at 21:56
jessicajessica
42
42
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that by definition,
$$
phi(x+y)=
begin{bmatrix}
cos(2(x+y)) & sin(2(x+y)) \
-sin(2(x+y)) & cos(2(x+y))
end{bmatrix}
$$
Now note that
begin{align}
cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
end{align}
and compute
$$
phi(x)phi(y)=
begin{bmatrix}
cos(2x) & sin(2x) \
-sin(2x) & cos(2x)
end{bmatrix}
begin{bmatrix}
cos(2y) & sin(2y) \
-sin(2y) & cos(2y)
end{bmatrix}
$$
as a matrix product.
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
$endgroup$
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024741%2fmatrix-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that by definition,
$$
phi(x+y)=
begin{bmatrix}
cos(2(x+y)) & sin(2(x+y)) \
-sin(2(x+y)) & cos(2(x+y))
end{bmatrix}
$$
Now note that
begin{align}
cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
end{align}
and compute
$$
phi(x)phi(y)=
begin{bmatrix}
cos(2x) & sin(2x) \
-sin(2x) & cos(2x)
end{bmatrix}
begin{bmatrix}
cos(2y) & sin(2y) \
-sin(2y) & cos(2y)
end{bmatrix}
$$
as a matrix product.
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
$endgroup$
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
add a comment |
$begingroup$
Note that by definition,
$$
phi(x+y)=
begin{bmatrix}
cos(2(x+y)) & sin(2(x+y)) \
-sin(2(x+y)) & cos(2(x+y))
end{bmatrix}
$$
Now note that
begin{align}
cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
end{align}
and compute
$$
phi(x)phi(y)=
begin{bmatrix}
cos(2x) & sin(2x) \
-sin(2x) & cos(2x)
end{bmatrix}
begin{bmatrix}
cos(2y) & sin(2y) \
-sin(2y) & cos(2y)
end{bmatrix}
$$
as a matrix product.
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
$endgroup$
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
add a comment |
$begingroup$
Note that by definition,
$$
phi(x+y)=
begin{bmatrix}
cos(2(x+y)) & sin(2(x+y)) \
-sin(2(x+y)) & cos(2(x+y))
end{bmatrix}
$$
Now note that
begin{align}
cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
end{align}
and compute
$$
phi(x)phi(y)=
begin{bmatrix}
cos(2x) & sin(2x) \
-sin(2x) & cos(2x)
end{bmatrix}
begin{bmatrix}
cos(2y) & sin(2y) \
-sin(2y) & cos(2y)
end{bmatrix}
$$
as a matrix product.
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
$endgroup$
Note that by definition,
$$
phi(x+y)=
begin{bmatrix}
cos(2(x+y)) & sin(2(x+y)) \
-sin(2(x+y)) & cos(2(x+y))
end{bmatrix}
$$
Now note that
begin{align}
cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
end{align}
and compute
$$
phi(x)phi(y)=
begin{bmatrix}
cos(2x) & sin(2x) \
-sin(2x) & cos(2x)
end{bmatrix}
begin{bmatrix}
cos(2y) & sin(2y) \
-sin(2y) & cos(2y)
end{bmatrix}
$$
as a matrix product.
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
edited Dec 4 '18 at 12:41
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
answered Dec 3 '18 at 22:14
egregegreg
183k1486204
183k1486204
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
add a comment |
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
$endgroup$
– jessica
Dec 4 '18 at 11:50
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
@jessica Why the determinant?
$endgroup$
– egreg
Dec 4 '18 at 12:37
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
:Don't worry I've found it. But can you help me with finding the kernel?
$endgroup$
– jessica
Dec 4 '18 at 12:54
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
$begingroup$
@jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
$endgroup$
– egreg
Dec 4 '18 at 13:02
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024741%2fmatrix-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
