Finding the matrix form of a linear operator with a non canonical vector space $V$?
$begingroup$
Given a vector space $V ={acos(t) + bsin(t) + ctsin(t)}$ with $a,b,c$ all real, a basis $B = {cos(t),sin(t),tsin(t)}$ and a Linear operator $Lf(t) = f''(t)$, how would I write ${lbrack Lrbrack}_{BB}$?
I attempted it and got begin{bmatrix}
-1 & 0 & 2t \
0 & -1 & 0 \
0 & 0 & -t
end{bmatrix}
however my professor shows the matrix representation as being
begin{bmatrix}
-1 & 0 & 2 \
0 & -1 & 0 \
0 & 0 & -1
end{bmatrix}
and so I am unsure as to why the $t$ disappeared
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 1:47
user3491700user3491700
535
$endgroup$
add a comment |
$begingroup$
Given a vector space $V ={acos(t) + bsin(t) + ctsin(t)}$ with $a,b,c$ all real, a basis $B = {cos(t),sin(t),tsin(t)}$ and a Linear operator $Lf(t) = f''(t)$, how would I write ${lbrack Lrbrack}_{BB}$?
I attempted it and got begin{bmatrix}
-1 & 0 & 2t \
0 & -1 & 0 \
0 & 0 & -t
end{bmatrix}
however my professor shows the matrix representation as being
begin{bmatrix}
-1 & 0 & 2 \
0 & -1 & 0 \
0 & 0 & -1
end{bmatrix}
and so I am unsure as to why the $t$ disappeared
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 1:47
user3491700user3491700
535
$endgroup$
1
$begingroup$
$L(tsin(t))=(tcos(t)+sin(t))’=tsin(t)+2cos(t)$. So L(third vector of B) is 2[first vector of B]+(-1)[last vector of B]
$endgroup$
– Mindlack
Dec 13 '18 at 1:52
$begingroup$
Well that's embarrassing, you're right, thanks
$endgroup$
– user3491700
Dec 13 '18 at 2:00
add a comment |
$begingroup$
Given a vector space $V ={acos(t) + bsin(t) + ctsin(t)}$ with $a,b,c$ all real, a basis $B = {cos(t),sin(t),tsin(t)}$ and a Linear operator $Lf(t) = f''(t)$, how would I write ${lbrack Lrbrack}_{BB}$?
I attempted it and got begin{bmatrix}
-1 & 0 & 2t \
0 & -1 & 0 \
0 & 0 & -t
end{bmatrix}
however my professor shows the matrix representation as being
begin{bmatrix}
-1 & 0 & 2 \
0 & -1 & 0 \
0 & 0 & -1
end{bmatrix}
and so I am unsure as to why the $t$ disappeared
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 1:47
user3491700user3491700
535
$endgroup$
Given a vector space $V ={acos(t) + bsin(t) + ctsin(t)}$ with $a,b,c$ all real, a basis $B = {cos(t),sin(t),tsin(t)}$ and a Linear operator $Lf(t) = f''(t)$, how would I write ${lbrack Lrbrack}_{BB}$?
I attempted it and got begin{bmatrix}
-1 & 0 & 2t \
0 & -1 & 0 \
0 & 0 & -t
end{bmatrix}
however my professor shows the matrix representation as being
begin{bmatrix}
-1 & 0 & 2 \
0 & -1 & 0 \
0 & 0 & -1
end{bmatrix}
and so I am unsure as to why the $t$ disappeared
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 1:47
user3491700user3491700
535
asked Dec 13 '18 at 1:47
user3491700user3491700
535
asked Dec 13 '18 at 1:47
user3491700user3491700
535
asked Dec 13 '18 at 1:47
user3491700user3491700
535
asked Dec 13 '18 at 1:47
user3491700user3491700
535
535
1
$begingroup$
$L(tsin(t))=(tcos(t)+sin(t))’=tsin(t)+2cos(t)$. So L(third vector of B) is 2[first vector of B]+(-1)[last vector of B]
$endgroup$
– Mindlack
Dec 13 '18 at 1:52
$begingroup$
Well that's embarrassing, you're right, thanks
$endgroup$
– user3491700
Dec 13 '18 at 2:00
add a comment |
1
$begingroup$
$L(tsin(t))=(tcos(t)+sin(t))’=tsin(t)+2cos(t)$. So L(third vector of B) is 2[first vector of B]+(-1)[last vector of B]
$endgroup$
– Mindlack
Dec 13 '18 at 1:52
$begingroup$
Well that's embarrassing, you're right, thanks
$endgroup$
– user3491700
Dec 13 '18 at 2:00
1
1
$begingroup$
$L(tsin(t))=(tcos(t)+sin(t))’=tsin(t)+2cos(t)$. So L(third vector of B) is 2[first vector of B]+(-1)[last vector of B]
$endgroup$
– Mindlack
Dec 13 '18 at 1:52
$begingroup$
$L(tsin(t))=(tcos(t)+sin(t))’=tsin(t)+2cos(t)$. So L(third vector of B) is 2[first vector of B]+(-1)[last vector of B]
$endgroup$
– Mindlack
Dec 13 '18 at 1:52
$begingroup$
Well that's embarrassing, you're right, thanks
$endgroup$
– user3491700
Dec 13 '18 at 2:00
$begingroup$
Well that's embarrassing, you're right, thanks
$endgroup$
– user3491700
Dec 13 '18 at 2:00
add a comment |
0
active
oldest
votes
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%2f3037496%2ffinding-the-matrix-form-of-a-linear-operator-with-a-non-canonical-vector-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3037496%2ffinding-the-matrix-form-of-a-linear-operator-with-a-non-canonical-vector-space%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
1
$begingroup$
$L(tsin(t))=(tcos(t)+sin(t))’=tsin(t)+2cos(t)$. So L(third vector of B) is 2[first vector of B]+(-1)[last vector of B]
$endgroup$
– Mindlack
Dec 13 '18 at 1:52
$begingroup$
Well that's embarrassing, you're right, thanks
$endgroup$
– user3491700
Dec 13 '18 at 2:00