Mixing
Proving that there is a unique linear map such that $T(u_i)=v_i$.
$begingroup$
I have a problem with understanding of a rather simple concept in linear algebra. I have seen in a book, a following question:
Suppose $U,V$ are vector spaces over $K$ and $u_1,dots,u_n$ is a basis of $U$. Let $v_1,dots, v_n$ be any sequence of $n$ vectors in $V$. Prove that there is a unique map $T:Urightarrow V$ with $T(u_i)=v_i$, for $1leq i leq n$.
The proof given in this book is:
Let $u in U$. Then since $u_1,dots, u_n$ is a basis of $U$, there exists uniquely determined $alpha_1,dots,alpha_n in K$ with $u=alpha_1 u_1+dots+alpha_n u_n$. So that, if $T$ exists, then we must have:
$$T(u)=T(alpha_1 u_1+dots+alpha_n u_n)=alpha_1 v_1+dots+alpha_n v_n$$
and so $T$ is uniquely determined.
Honestly this proof makes little sense to me. How does it prove uniqueness? Is this proof complete? Could someone explain this to me? Thank you very much.
linear-algebra vector-spaces proof-writing
asked May 11 '15 at 20:49
marco11marco11
489317
$endgroup$
add a comment |
$begingroup$
I have a problem with understanding of a rather simple concept in linear algebra. I have seen in a book, a following question:
Suppose $U,V$ are vector spaces over $K$ and $u_1,dots,u_n$ is a basis of $U$. Let $v_1,dots, v_n$ be any sequence of $n$ vectors in $V$. Prove that there is a unique map $T:Urightarrow V$ with $T(u_i)=v_i$, for $1leq i leq n$.
The proof given in this book is:
Let $u in U$. Then since $u_1,dots, u_n$ is a basis of $U$, there exists uniquely determined $alpha_1,dots,alpha_n in K$ with $u=alpha_1 u_1+dots+alpha_n u_n$. So that, if $T$ exists, then we must have:
$$T(u)=T(alpha_1 u_1+dots+alpha_n u_n)=alpha_1 v_1+dots+alpha_n v_n$$
and so $T$ is uniquely determined.
Honestly this proof makes little sense to me. How does it prove uniqueness? Is this proof complete? Could someone explain this to me? Thank you very much.
linear-algebra vector-spaces proof-writing
asked May 11 '15 at 20:49
marco11marco11
489317
$endgroup$
3
$begingroup$
Repeat the argument for 'another' map, say $T'$. You get $T(u)=T'(u)$, for all $uin U$. Edit: An analyst would say that the RHS, $alpha_1 v_1+dots+alpha_n v_n$, does not depend on $T$, so all such maps will do the same, hence uniqueness.
$endgroup$
– Git Gud
May 11 '15 at 20:54
1
$begingroup$
It is just linearity; if you asked about another T you have to define it otherwise
$endgroup$
– Piquito
May 11 '15 at 20:57
$begingroup$
I still don't really get this. If a take another map, why the RHS does not change?
$endgroup$
– marco11
May 11 '15 at 21:03
add a comment |
$begingroup$
I have a problem with understanding of a rather simple concept in linear algebra. I have seen in a book, a following question:
Suppose $U,V$ are vector spaces over $K$ and $u_1,dots,u_n$ is a basis of $U$. Let $v_1,dots, v_n$ be any sequence of $n$ vectors in $V$. Prove that there is a unique map $T:Urightarrow V$ with $T(u_i)=v_i$, for $1leq i leq n$.
The proof given in this book is:
Let $u in U$. Then since $u_1,dots, u_n$ is a basis of $U$, there exists uniquely determined $alpha_1,dots,alpha_n in K$ with $u=alpha_1 u_1+dots+alpha_n u_n$. So that, if $T$ exists, then we must have:
$$T(u)=T(alpha_1 u_1+dots+alpha_n u_n)=alpha_1 v_1+dots+alpha_n v_n$$
and so $T$ is uniquely determined.
Honestly this proof makes little sense to me. How does it prove uniqueness? Is this proof complete? Could someone explain this to me? Thank you very much.
linear-algebra vector-spaces proof-writing
asked May 11 '15 at 20:49
marco11marco11
489317
$endgroup$
I have a problem with understanding of a rather simple concept in linear algebra. I have seen in a book, a following question:
Suppose $U,V$ are vector spaces over $K$ and $u_1,dots,u_n$ is a basis of $U$. Let $v_1,dots, v_n$ be any sequence of $n$ vectors in $V$. Prove that there is a unique map $T:Urightarrow V$ with $T(u_i)=v_i$, for $1leq i leq n$.
The proof given in this book is:
Let $u in U$. Then since $u_1,dots, u_n$ is a basis of $U$, there exists uniquely determined $alpha_1,dots,alpha_n in K$ with $u=alpha_1 u_1+dots+alpha_n u_n$. So that, if $T$ exists, then we must have:
$$T(u)=T(alpha_1 u_1+dots+alpha_n u_n)=alpha_1 v_1+dots+alpha_n v_n$$
and so $T$ is uniquely determined.
Honestly this proof makes little sense to me. How does it prove uniqueness? Is this proof complete? Could someone explain this to me? Thank you very much.
linear-algebra vector-spaces proof-writing
linear-algebra vector-spaces proof-writing
asked May 11 '15 at 20:49
marco11marco11
489317
asked May 11 '15 at 20:49
marco11marco11
489317
asked May 11 '15 at 20:49
marco11marco11
489317
asked May 11 '15 at 20:49
marco11marco11
489317
asked May 11 '15 at 20:49
marco11marco11
489317
489317
3
$begingroup$
Repeat the argument for 'another' map, say $T'$. You get $T(u)=T'(u)$, for all $uin U$. Edit: An analyst would say that the RHS, $alpha_1 v_1+dots+alpha_n v_n$, does not depend on $T$, so all such maps will do the same, hence uniqueness.
$endgroup$
– Git Gud
May 11 '15 at 20:54
1
$begingroup$
It is just linearity; if you asked about another T you have to define it otherwise
$endgroup$
– Piquito
May 11 '15 at 20:57
$begingroup$
I still don't really get this. If a take another map, why the RHS does not change?
$endgroup$
– marco11
May 11 '15 at 21:03
add a comment |
3
$begingroup$
Repeat the argument for 'another' map, say $T'$. You get $T(u)=T'(u)$, for all $uin U$. Edit: An analyst would say that the RHS, $alpha_1 v_1+dots+alpha_n v_n$, does not depend on $T$, so all such maps will do the same, hence uniqueness.
$endgroup$
– Git Gud
May 11 '15 at 20:54
1
$begingroup$
It is just linearity; if you asked about another T you have to define it otherwise
$endgroup$
– Piquito
May 11 '15 at 20:57
$begingroup$
I still don't really get this. If a take another map, why the RHS does not change?
$endgroup$
– marco11
May 11 '15 at 21:03
3
3
$begingroup$
Repeat the argument for 'another' map, say $T'$. You get $T(u)=T'(u)$, for all $uin U$. Edit: An analyst would say that the RHS, $alpha_1 v_1+dots+alpha_n v_n$, does not depend on $T$, so all such maps will do the same, hence uniqueness.
$endgroup$
– Git Gud
May 11 '15 at 20:54
$begingroup$
Repeat the argument for 'another' map, say $T'$. You get $T(u)=T'(u)$, for all $uin U$. Edit: An analyst would say that the RHS, $alpha_1 v_1+dots+alpha_n v_n$, does not depend on $T$, so all such maps will do the same, hence uniqueness.
$endgroup$
– Git Gud
May 11 '15 at 20:54
1
1
$begingroup$
It is just linearity; if you asked about another T you have to define it otherwise
$endgroup$
– Piquito
May 11 '15 at 20:57
$begingroup$
It is just linearity; if you asked about another T you have to define it otherwise
$endgroup$
– Piquito
May 11 '15 at 20:57
$begingroup$
I still don't really get this. If a take another map, why the RHS does not change?
$endgroup$
– marco11
May 11 '15 at 21:03
$begingroup$
I still don't really get this. If a take another map, why the RHS does not change?
$endgroup$
– marco11
May 11 '15 at 21:03
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let's say we have another mapping $f:U to V$ such that $f(u_i)=v_i, forall i$. Then, for each $u in U:$ $T(u)=T(a_1u_1+dots + a_nu_n)=a_1T(u_i)+dots + a_nT(u_n)=a_1v_1+ dots a_nv_n=a_1f(u_1)+ dots + a_nf(u_n)$
$=f(a_1u_1+ dots +a_nu_n)=f(u)$
Hope that answers your question.
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
$endgroup$
add a comment |
$begingroup$
You like for the map $T$ to be a function of $u$ and $v_1,v_2,cdots, v_n$. You can do so by making a slight change of notation $u = displaystyle sum_{i=1}^n alpha_iu_i = (alpha_1,alpha_2,cdots,alpha_n)$, and $A = begin{pmatrix} v_1\v_2\...\v_nend{pmatrix}$, then define $T:U to V: T(u) = ucdot A$, then you check that $T(u_i) = (0,0,cdots,1,cdots,0)cdot begin{pmatrix} v_1\ v_2\...\v_i\...\v_nend{pmatrix}=v_i$. Proving uniqueness is simple for if there is another map $T': U to V: T'(u_i)=v_i to T'(u) = ucdot A = T(u) to T'=T$.
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
$endgroup$
add a comment |
$begingroup$
note that a linear map is completely determined by its values on basis of U. Specifically from the T(u)=T(α1u1+⋯+αnun)=α1v1+⋯+αnvn, you could see how the result of the linear map is merely the linear combination of the mapping of basis on V. (example given by DeepSea) So the question becomes does the value on the basis determines a T. Of course yes! Since the linear combination of the basis of U covers its vector space, so it determines mapping result of each element in U. Then the question is whether the mapping relationship between U and V could be represented by multiple linear maps on the same basis? (it can not. see the answer by cold coffee)
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
$endgroup$
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's say we have another mapping $f:U to V$ such that $f(u_i)=v_i, forall i$. Then, for each $u in U:$ $T(u)=T(a_1u_1+dots + a_nu_n)=a_1T(u_i)+dots + a_nT(u_n)=a_1v_1+ dots a_nv_n=a_1f(u_1)+ dots + a_nf(u_n)$
$=f(a_1u_1+ dots +a_nu_n)=f(u)$
Hope that answers your question.
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
$endgroup$
add a comment |
$begingroup$
Let's say we have another mapping $f:U to V$ such that $f(u_i)=v_i, forall i$. Then, for each $u in U:$ $T(u)=T(a_1u_1+dots + a_nu_n)=a_1T(u_i)+dots + a_nT(u_n)=a_1v_1+ dots a_nv_n=a_1f(u_1)+ dots + a_nf(u_n)$
$=f(a_1u_1+ dots +a_nu_n)=f(u)$
Hope that answers your question.
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
$endgroup$
add a comment |
$begingroup$
Let's say we have another mapping $f:U to V$ such that $f(u_i)=v_i, forall i$. Then, for each $u in U:$ $T(u)=T(a_1u_1+dots + a_nu_n)=a_1T(u_i)+dots + a_nT(u_n)=a_1v_1+ dots a_nv_n=a_1f(u_1)+ dots + a_nf(u_n)$
$=f(a_1u_1+ dots +a_nu_n)=f(u)$
Hope that answers your question.
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
$endgroup$
Let's say we have another mapping $f:U to V$ such that $f(u_i)=v_i, forall i$. Then, for each $u in U:$ $T(u)=T(a_1u_1+dots + a_nu_n)=a_1T(u_i)+dots + a_nT(u_n)=a_1v_1+ dots a_nv_n=a_1f(u_1)+ dots + a_nf(u_n)$
$=f(a_1u_1+ dots +a_nu_n)=f(u)$
Hope that answers your question.
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
answered May 11 '15 at 21:19
coldcoffeecoldcoffee
1097
1097
add a comment |
add a comment |
$begingroup$
You like for the map $T$ to be a function of $u$ and $v_1,v_2,cdots, v_n$. You can do so by making a slight change of notation $u = displaystyle sum_{i=1}^n alpha_iu_i = (alpha_1,alpha_2,cdots,alpha_n)$, and $A = begin{pmatrix} v_1\v_2\...\v_nend{pmatrix}$, then define $T:U to V: T(u) = ucdot A$, then you check that $T(u_i) = (0,0,cdots,1,cdots,0)cdot begin{pmatrix} v_1\ v_2\...\v_i\...\v_nend{pmatrix}=v_i$. Proving uniqueness is simple for if there is another map $T': U to V: T'(u_i)=v_i to T'(u) = ucdot A = T(u) to T'=T$.
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
$endgroup$
add a comment |
$begingroup$
You like for the map $T$ to be a function of $u$ and $v_1,v_2,cdots, v_n$. You can do so by making a slight change of notation $u = displaystyle sum_{i=1}^n alpha_iu_i = (alpha_1,alpha_2,cdots,alpha_n)$, and $A = begin{pmatrix} v_1\v_2\...\v_nend{pmatrix}$, then define $T:U to V: T(u) = ucdot A$, then you check that $T(u_i) = (0,0,cdots,1,cdots,0)cdot begin{pmatrix} v_1\ v_2\...\v_i\...\v_nend{pmatrix}=v_i$. Proving uniqueness is simple for if there is another map $T': U to V: T'(u_i)=v_i to T'(u) = ucdot A = T(u) to T'=T$.
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
$endgroup$
add a comment |
$begingroup$
You like for the map $T$ to be a function of $u$ and $v_1,v_2,cdots, v_n$. You can do so by making a slight change of notation $u = displaystyle sum_{i=1}^n alpha_iu_i = (alpha_1,alpha_2,cdots,alpha_n)$, and $A = begin{pmatrix} v_1\v_2\...\v_nend{pmatrix}$, then define $T:U to V: T(u) = ucdot A$, then you check that $T(u_i) = (0,0,cdots,1,cdots,0)cdot begin{pmatrix} v_1\ v_2\...\v_i\...\v_nend{pmatrix}=v_i$. Proving uniqueness is simple for if there is another map $T': U to V: T'(u_i)=v_i to T'(u) = ucdot A = T(u) to T'=T$.
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
$endgroup$
You like for the map $T$ to be a function of $u$ and $v_1,v_2,cdots, v_n$. You can do so by making a slight change of notation $u = displaystyle sum_{i=1}^n alpha_iu_i = (alpha_1,alpha_2,cdots,alpha_n)$, and $A = begin{pmatrix} v_1\v_2\...\v_nend{pmatrix}$, then define $T:U to V: T(u) = ucdot A$, then you check that $T(u_i) = (0,0,cdots,1,cdots,0)cdot begin{pmatrix} v_1\ v_2\...\v_i\...\v_nend{pmatrix}=v_i$. Proving uniqueness is simple for if there is another map $T': U to V: T'(u_i)=v_i to T'(u) = ucdot A = T(u) to T'=T$.
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
answered May 11 '15 at 21:26
DeepSeaDeepSea
71.2k54487
71.2k54487
add a comment |
add a comment |
$begingroup$
note that a linear map is completely determined by its values on basis of U. Specifically from the T(u)=T(α1u1+⋯+αnun)=α1v1+⋯+αnvn, you could see how the result of the linear map is merely the linear combination of the mapping of basis on V. (example given by DeepSea) So the question becomes does the value on the basis determines a T. Of course yes! Since the linear combination of the basis of U covers its vector space, so it determines mapping result of each element in U. Then the question is whether the mapping relationship between U and V could be represented by multiple linear maps on the same basis? (it can not. see the answer by cold coffee)
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
$endgroup$
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
add a comment |
$begingroup$
note that a linear map is completely determined by its values on basis of U. Specifically from the T(u)=T(α1u1+⋯+αnun)=α1v1+⋯+αnvn, you could see how the result of the linear map is merely the linear combination of the mapping of basis on V. (example given by DeepSea) So the question becomes does the value on the basis determines a T. Of course yes! Since the linear combination of the basis of U covers its vector space, so it determines mapping result of each element in U. Then the question is whether the mapping relationship between U and V could be represented by multiple linear maps on the same basis? (it can not. see the answer by cold coffee)
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
$endgroup$
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
add a comment |
$begingroup$
note that a linear map is completely determined by its values on basis of U. Specifically from the T(u)=T(α1u1+⋯+αnun)=α1v1+⋯+αnvn, you could see how the result of the linear map is merely the linear combination of the mapping of basis on V. (example given by DeepSea) So the question becomes does the value on the basis determines a T. Of course yes! Since the linear combination of the basis of U covers its vector space, so it determines mapping result of each element in U. Then the question is whether the mapping relationship between U and V could be represented by multiple linear maps on the same basis? (it can not. see the answer by cold coffee)
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
$endgroup$
note that a linear map is completely determined by its values on basis of U. Specifically from the T(u)=T(α1u1+⋯+αnun)=α1v1+⋯+αnvn, you could see how the result of the linear map is merely the linear combination of the mapping of basis on V. (example given by DeepSea) So the question becomes does the value on the basis determines a T. Of course yes! Since the linear combination of the basis of U covers its vector space, so it determines mapping result of each element in U. Then the question is whether the mapping relationship between U and V could be represented by multiple linear maps on the same basis? (it can not. see the answer by cold coffee)
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
answered Jan 9 at 11:23
Sitan LiuSitan Liu
1
1
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
add a comment |
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
I don't see what this adds to the existing answers (except being harder to read). Please don't waste your time writing redundant answers to old questions, nobody benefits from that.
$endgroup$
– Henrik
Jan 9 at 11:43
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
$begingroup$
Thanks for your comment I will be more careful in the future.
$endgroup$
– Sitan Liu
Jan 10 at 5:28
add a comment |
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Repeat the argument for 'another' map, say $T'$. You get $T(u)=T'(u)$, for all $uin U$. Edit: An analyst would say that the RHS, $alpha_1 v_1+dots+alpha_n v_n$, does not depend on $T$, so all such maps will do the same, hence uniqueness.
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– Git Gud
May 11 '15 at 20:54
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It is just linearity; if you asked about another T you have to define it otherwise
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– Piquito
May 11 '15 at 20:57
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I still don't really get this. If a take another map, why the RHS does not change?
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– marco11
May 11 '15 at 21:03