Why is the Change of Basis map unique?
$begingroup$
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, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$, 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
4,0611320
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
$endgroup$
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, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$, 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
4,0611320
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
$endgroup$
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, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$, 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
4,0611320
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
$endgroup$
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, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$, 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
4,0611320
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
edited Feb 8 at 1:19
J. W. Tanner
4,0611320
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
edited Feb 8 at 1:19
J. W. Tanner
4,0611320
edited Feb 8 at 1:19
J. W. Tanner
4,0611320
edited Feb 8 at 1:19
J. W. Tanner
4,0611320
4,0611320
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
asked Feb 7 at 6:50
Joe Man AnalysisJoe Man Analysis
56619
56619
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
180k1494233
$endgroup$
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
1,158114
$endgroup$
add a comment |
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2 Answers
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2 Answers
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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
180k1494233
$endgroup$
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
180k1494233
$endgroup$
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
180k1494233
$endgroup$
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
180k1494233
answered Feb 7 at 6:56
DonAntonioDonAntonio
180k1494233
answered Feb 7 at 6:56
DonAntonioDonAntonio
180k1494233
answered Feb 7 at 6:56
DonAntonioDonAntonio
180k1494233
180k1494233
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
1,158114
$endgroup$
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
1,158114
$endgroup$
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
1,158114
$endgroup$
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
1,158114
answered Feb 7 at 6:57
AlessioDVAlessioDV
1,158114
answered Feb 7 at 6:57
AlessioDVAlessioDV
1,158114
answered Feb 7 at 6:57
AlessioDVAlessioDV
1,158114
1,158114
add a comment |
add a comment |
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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