Counterclockwise rotation matrix
$begingroup$
If I take the basis $(vec{e_x},vec{e_y})$ and make a rotation counterclockwise of angle $theta$, I end up with two new vectors $(vec{u},vec{v})$ such that :
$vec{u} = costheta vec{e_x} + sintheta vec{e_y}$
$vec{v} = costheta vec{e_x} - sintheta vec{e_y}$
so
begin{equation}
left( begin{array}{ccc}
vec{u} \
vec{v}end{array} right)
= left( begin{array}{ccc}
costheta & sintheta\
-sintheta & costhetaend{array} right)
left( begin{array}{ccc}
vec{e_x} \
vec{e_y}end{array} right)
end{equation}
I don't understand why the counterclockwise rotation is defined as :
begin{pmatrix}
costheta & -sintheta \
sintheta & costheta
end{pmatrix}
EDIT:
When I look at my picture, it looks like a counterclockwise rotation...
matrices rotations
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
asked May 11 '15 at 8:40
user1234161user1234161
290310
$endgroup$
add a comment |
$begingroup$
If I take the basis $(vec{e_x},vec{e_y})$ and make a rotation counterclockwise of angle $theta$, I end up with two new vectors $(vec{u},vec{v})$ such that :
$vec{u} = costheta vec{e_x} + sintheta vec{e_y}$
$vec{v} = costheta vec{e_x} - sintheta vec{e_y}$
so
begin{equation}
left( begin{array}{ccc}
vec{u} \
vec{v}end{array} right)
= left( begin{array}{ccc}
costheta & sintheta\
-sintheta & costhetaend{array} right)
left( begin{array}{ccc}
vec{e_x} \
vec{e_y}end{array} right)
end{equation}
I don't understand why the counterclockwise rotation is defined as :
begin{pmatrix}
costheta & -sintheta \
sintheta & costheta
end{pmatrix}
EDIT:
When I look at my picture, it looks like a counterclockwise rotation...
matrices rotations
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
asked May 11 '15 at 8:40
user1234161user1234161
290310
$endgroup$
add a comment |
$begingroup$
If I take the basis $(vec{e_x},vec{e_y})$ and make a rotation counterclockwise of angle $theta$, I end up with two new vectors $(vec{u},vec{v})$ such that :
$vec{u} = costheta vec{e_x} + sintheta vec{e_y}$
$vec{v} = costheta vec{e_x} - sintheta vec{e_y}$
so
begin{equation}
left( begin{array}{ccc}
vec{u} \
vec{v}end{array} right)
= left( begin{array}{ccc}
costheta & sintheta\
-sintheta & costhetaend{array} right)
left( begin{array}{ccc}
vec{e_x} \
vec{e_y}end{array} right)
end{equation}
I don't understand why the counterclockwise rotation is defined as :
begin{pmatrix}
costheta & -sintheta \
sintheta & costheta
end{pmatrix}
EDIT:
When I look at my picture, it looks like a counterclockwise rotation...
matrices rotations
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
asked May 11 '15 at 8:40
user1234161user1234161
290310
$endgroup$
If I take the basis $(vec{e_x},vec{e_y})$ and make a rotation counterclockwise of angle $theta$, I end up with two new vectors $(vec{u},vec{v})$ such that :
$vec{u} = costheta vec{e_x} + sintheta vec{e_y}$
$vec{v} = costheta vec{e_x} - sintheta vec{e_y}$
so
begin{equation}
left( begin{array}{ccc}
vec{u} \
vec{v}end{array} right)
= left( begin{array}{ccc}
costheta & sintheta\
-sintheta & costhetaend{array} right)
left( begin{array}{ccc}
vec{e_x} \
vec{e_y}end{array} right)
end{equation}
I don't understand why the counterclockwise rotation is defined as :
begin{pmatrix}
costheta & -sintheta \
sintheta & costheta
end{pmatrix}
EDIT:
When I look at my picture, it looks like a counterclockwise rotation...
matrices rotations
matrices rotations
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
asked May 11 '15 at 8:40
user1234161user1234161
290310
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
asked May 11 '15 at 8:40
user1234161user1234161
290310
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
edited May 11 '15 at 11:01
Martin Sleziak
44.7k9117272
44.7k9117272
asked May 11 '15 at 8:40
user1234161user1234161
290310
asked May 11 '15 at 8:40
user1234161user1234161
290310
asked May 11 '15 at 8:40
user1234161user1234161
290310
290310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose the rotation matrix is
$$begin{bmatrix}a&b\c&dend{bmatrix}$$
Since it rotate every vector by angle $theta$, we will look at what it does to the basis $begin{bmatrix}1\0end{bmatrix}$, $begin{bmatrix}0\1end{bmatrix}$.
$$begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}1\0end{bmatrix}=begin{bmatrix}a\cend{bmatrix}$$
By the following picture, we could see that $a=costheta,c=sintheta$.
Similarly, you can find $b,d$.
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
$endgroup$
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
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%2f1276814%2fcounterclockwise-rotation-matrix%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
$begingroup$
Suppose the rotation matrix is
$$begin{bmatrix}a&b\c&dend{bmatrix}$$
Since it rotate every vector by angle $theta$, we will look at what it does to the basis $begin{bmatrix}1\0end{bmatrix}$, $begin{bmatrix}0\1end{bmatrix}$.
$$begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}1\0end{bmatrix}=begin{bmatrix}a\cend{bmatrix}$$
By the following picture, we could see that $a=costheta,c=sintheta$.
Similarly, you can find $b,d$.
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
$endgroup$
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
add a comment |
$begingroup$
Suppose the rotation matrix is
$$begin{bmatrix}a&b\c&dend{bmatrix}$$
Since it rotate every vector by angle $theta$, we will look at what it does to the basis $begin{bmatrix}1\0end{bmatrix}$, $begin{bmatrix}0\1end{bmatrix}$.
$$begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}1\0end{bmatrix}=begin{bmatrix}a\cend{bmatrix}$$
By the following picture, we could see that $a=costheta,c=sintheta$.
Similarly, you can find $b,d$.
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
$endgroup$
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
add a comment |
$begingroup$
Suppose the rotation matrix is
$$begin{bmatrix}a&b\c&dend{bmatrix}$$
Since it rotate every vector by angle $theta$, we will look at what it does to the basis $begin{bmatrix}1\0end{bmatrix}$, $begin{bmatrix}0\1end{bmatrix}$.
$$begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}1\0end{bmatrix}=begin{bmatrix}a\cend{bmatrix}$$
By the following picture, we could see that $a=costheta,c=sintheta$.
Similarly, you can find $b,d$.
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
$endgroup$
Suppose the rotation matrix is
$$begin{bmatrix}a&b\c&dend{bmatrix}$$
Since it rotate every vector by angle $theta$, we will look at what it does to the basis $begin{bmatrix}1\0end{bmatrix}$, $begin{bmatrix}0\1end{bmatrix}$.
$$begin{bmatrix}a&b\c&dend{bmatrix}begin{bmatrix}1\0end{bmatrix}=begin{bmatrix}a\cend{bmatrix}$$
By the following picture, we could see that $a=costheta,c=sintheta$.
Similarly, you can find $b,d$.
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
answered May 11 '15 at 9:12
KittyLKittyL
13.8k31534
13.8k31534
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
add a comment |
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
But what is wrong with what I did ?
$endgroup$
– user1234161
May 11 '15 at 9:33
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
$begingroup$
@user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that.
$endgroup$
– KittyL
May 11 '15 at 9:36
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%2f1276814%2fcounterclockwise-rotation-matrix%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
