Why is the Change of Basis map unique?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I've been looking all over, but I haven't found anything satisfactory.
We've been shown in class by a commutative diagram that, given an $n$-dimensional vector space $V$ over a field, $mathbbF$, and bases, $mathcalB=v_1,...,v_n$ and $mathcalC=u_1,...,u_n$, the coordinate maps $_mathcalB:Vrightarrow mathbbF^n$ and $_mathcalC:Vrightarrow mathbbF^n$ give rise to a unique map $P=_mathcalBcirc ^-1_mathcalC:mathbbF^nrightarrow mathbbF^n$, which is our change of basis matrix.
But I am having a lot of trouble proving that $P$ is unique. Can anyone enlighten me as to why this is necessarily true?
linear-algebra linear-transformations
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
$endgroup$
add a comment |
$begingroup$
I've been looking all over, but I haven't found anything satisfactory.
We've been shown in class by a commutative diagram that, given an $n$-dimensional vector space $V$ over a field, $mathbbF$, and bases, $mathcalB=v_1,...,v_n$ and $mathcalC=u_1,...,u_n$, the coordinate maps $_mathcalB:Vrightarrow mathbbF^n$ and $_mathcalC:Vrightarrow mathbbF^n$ give rise to a unique map $P=_mathcalBcirc ^-1_mathcalC:mathbbF^nrightarrow mathbbF^n$, which is our change of basis matrix.
But I am having a lot of trouble proving that $P$ is unique. Can anyone enlighten me as to why this is necessarily true?
linear-algebra linear-transformations
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
$endgroup$
5
$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^-1$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22
$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42
add a comment |
$begingroup$
I've been looking all over, but I haven't found anything satisfactory.
We've been shown in class by a commutative diagram that, given an $n$-dimensional vector space $V$ over a field, $mathbbF$, and bases, $mathcalB=v_1,...,v_n$ and $mathcalC=u_1,...,u_n$, the coordinate maps $_mathcalB:Vrightarrow mathbbF^n$ and $_mathcalC:Vrightarrow mathbbF^n$ give rise to a unique map $P=_mathcalBcirc ^-1_mathcalC:mathbbF^nrightarrow mathbbF^n$, which is our change of basis matrix.
But I am having a lot of trouble proving that $P$ is unique. Can anyone enlighten me as to why this is necessarily true?
linear-algebra linear-transformations
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
$endgroup$
I've been looking all over, but I haven't found anything satisfactory.
We've been shown in class by a commutative diagram that, given an $n$-dimensional vector space $V$ over a field, $mathbbF$, and bases, $mathcalB=v_1,...,v_n$ and $mathcalC=u_1,...,u_n$, the coordinate maps $_mathcalB:Vrightarrow mathbbF^n$ and $_mathcalC:Vrightarrow mathbbF^n$ give rise to a unique map $P=_mathcalBcirc ^-1_mathcalC:mathbbF^nrightarrow mathbbF^n$, which is our change of basis matrix.
But I am having a lot of trouble proving that $P$ is unique. Can anyone enlighten me as to why this is necessarily true?
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
edited Feb 8 at 1:19
J. W. Tanner
2,6341217
2,6341217
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
50119
50119
5
$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^-1$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22
$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42
add a comment |
5
$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^-1$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22
$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42
5
5
$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^-1$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22
$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^-1$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22
$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42
$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.
If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any
$$v=sum_k=1^n a_iv_iin V;,;;Qv=sum_k=1^na_iQv_i=sum_k=1^n a_iPv_i=Pv$$
so $;Qequiv P;$.
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
$endgroup$
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%2f3103497%2fwhy-is-the-change-of-basis-map-unique%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.
If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any
$$v=sum_k=1^n a_iv_iin V;,;;Qv=sum_k=1^na_iQv_i=sum_k=1^n a_iPv_i=Pv$$
so $;Qequiv P;$.
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.
If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any
$$v=sum_k=1^n a_iv_iin V;,;;Qv=sum_k=1^na_iQv_i=sum_k=1^n a_iPv_i=Pv$$
so $;Qequiv P;$.
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.
If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any
$$v=sum_k=1^n a_iv_iin V;,;;Qv=sum_k=1^na_iQv_i=sum_k=1^n a_iPv_i=Pv$$
so $;Qequiv P;$.
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
$endgroup$
Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.
If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any
$$v=sum_k=1^n a_iv_iin V;,;;Qv=sum_k=1^na_iQv_i=sum_k=1^n a_iPv_i=Pv$$
so $;Qequiv P;$.
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
answered Feb 7 at 6:56
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
add a comment |
$begingroup$
A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
$endgroup$
add a comment |
$begingroup$
A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
$endgroup$
add a comment |
$begingroup$
A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
$endgroup$
A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
answered Feb 7 at 6:57
AlessioDVAlessioDV
655114
655114
add a comment |
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%2f3103497%2fwhy-is-the-change-of-basis-map-unique%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
5
$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^-1$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22
$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42