An easy proof that an isometry preserving the zero vector is linearWhat is a general scalar and what a (complex conjugate)Show that $|lambda|leq 1$ for each eigenvalue $lambda$ of a partial isometryKinds of finite dimensional inner product spacesconformal maps between vector spaces?$B$ be an uncountable dimensional real Banach space and $T:B to B$ be a surjective isometry such that $T(0)=0$ , then is $T$ linear?Limitations on transforming a vector into another vector with norm-preserving linear transformationsIs a non-euclidean-norm preserving map necessarily linear?What properties of a linear map can be determined from its matrix?Extension of a linear map in a generic vector space (without Zorn's lemma)Are all isometries of subsets affine?
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An easy proof that an isometry preserving the zero vector is linear
What is a general scalar and what a (complex conjugate)Show that $|lambda|leq 1$ for each eigenvalue $lambda$ of a partial isometryKinds of finite dimensional inner product spacesconformal maps between vector spaces?$B$ be an uncountable dimensional real Banach space and $T:B to B$ be a surjective isometry such that $T(0)=0$ , then is $T$ linear?Limitations on transforming a vector into another vector with norm-preserving linear transformationsIs a non-euclidean-norm preserving map necessarily linear?What properties of a linear map can be determined from its matrix?Extension of a linear map in a generic vector space (without Zorn's lemma)Are all isometries of subsets affine?
$begingroup$
I want to show that for real inner product spaces $V$ and $W$, if $L:Vto W$ satisfies the following properties:$$parallel L(vecx)-L(vecy)parallel=parallel vecx -vecyparallel\$$and $$L(vec0)=vec0,$$
then this map is linear. I am aware of the existence of the (more general) theorem of Mazur-Ulam, but I was wondering if there is more accessible proof, which is suitable for beginners in linear algebra. Thanks in advance!
linear-algebra linear-transformations isometry
asked 8 hours ago
EBPEBP
1786
$endgroup$
add a comment |
$begingroup$
I want to show that for real inner product spaces $V$ and $W$, if $L:Vto W$ satisfies the following properties:$$parallel L(vecx)-L(vecy)parallel=parallel vecx -vecyparallel\$$and $$L(vec0)=vec0,$$
then this map is linear. I am aware of the existence of the (more general) theorem of Mazur-Ulam, but I was wondering if there is more accessible proof, which is suitable for beginners in linear algebra. Thanks in advance!
linear-algebra linear-transformations isometry
asked 8 hours ago
EBPEBP
1786
$endgroup$
1
$begingroup$
Is the scalar field $mathbbR$ or $mathbbC$?
$endgroup$
– user159517
8 hours ago
$begingroup$
Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks.
$endgroup$
– EBP
8 hours ago
$begingroup$
Are we also assuming that $L$ is surjective?
$endgroup$
– user159517
8 hours ago
$begingroup$
No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity.
$endgroup$
– EBP
8 hours ago
add a comment |
$begingroup$
I want to show that for real inner product spaces $V$ and $W$, if $L:Vto W$ satisfies the following properties:$$parallel L(vecx)-L(vecy)parallel=parallel vecx -vecyparallel\$$and $$L(vec0)=vec0,$$
then this map is linear. I am aware of the existence of the (more general) theorem of Mazur-Ulam, but I was wondering if there is more accessible proof, which is suitable for beginners in linear algebra. Thanks in advance!
linear-algebra linear-transformations isometry
asked 8 hours ago
EBPEBP
1786
$endgroup$
I want to show that for real inner product spaces $V$ and $W$, if $L:Vto W$ satisfies the following properties:$$parallel L(vecx)-L(vecy)parallel=parallel vecx -vecyparallel\$$and $$L(vec0)=vec0,$$
then this map is linear. I am aware of the existence of the (more general) theorem of Mazur-Ulam, but I was wondering if there is more accessible proof, which is suitable for beginners in linear algebra. Thanks in advance!
linear-algebra linear-transformations isometry
linear-algebra linear-transformations isometry
asked 8 hours ago
EBPEBP
1786
asked 8 hours ago
EBPEBP
1786
edited 7 hours ago
EBP
asked 8 hours ago
EBPEBP
1786
asked 8 hours ago
EBPEBP
1786
asked 8 hours ago
EBPEBP
1786
1786
1
$begingroup$
Is the scalar field $mathbbR$ or $mathbbC$?
$endgroup$
– user159517
8 hours ago
$begingroup$
Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks.
$endgroup$
– EBP
8 hours ago
$begingroup$
Are we also assuming that $L$ is surjective?
$endgroup$
– user159517
8 hours ago
$begingroup$
No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity.
$endgroup$
– EBP
8 hours ago
add a comment |
1
$begingroup$
Is the scalar field $mathbbR$ or $mathbbC$?
$endgroup$
– user159517
8 hours ago
$begingroup$
Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks.
$endgroup$
– EBP
8 hours ago
$begingroup$
Are we also assuming that $L$ is surjective?
$endgroup$
– user159517
8 hours ago
$begingroup$
No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity.
$endgroup$
– EBP
8 hours ago
1
1
$begingroup$
Is the scalar field $mathbbR$ or $mathbbC$?
$endgroup$
– user159517
8 hours ago
$begingroup$
Is the scalar field $mathbbR$ or $mathbbC$?
$endgroup$
– user159517
8 hours ago
$begingroup$
Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks.
$endgroup$
– EBP
8 hours ago
$begingroup$
Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks.
$endgroup$
– EBP
8 hours ago
$begingroup$
Are we also assuming that $L$ is surjective?
$endgroup$
– user159517
8 hours ago
$begingroup$
Are we also assuming that $L$ is surjective?
$endgroup$
– user159517
8 hours ago
$begingroup$
No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity.
$endgroup$
– EBP
8 hours ago
$begingroup$
No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity.
$endgroup$
– EBP
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that the assumptions imply that if $L(x)=0$, then $x=0$. Besides, $|L(x)|=|x|$ for all $x$.
Now, note that $|a+b|=sqrt^2+$.
As a consequence, for all $x,y$, $langle L(x),, L(y) rangle=langle x,y rangle$.
Therefore, for any scalars $lambda_i$, for any vectors $x_i$, $$|sum_ilambda_iL(x_i)|=|sum_ilambda_ix_i|.$$
Then take $x_1=alpha u+beta v$, $x_2=u$, $x_3=v$, $lambda_1=-1$, $lambda_2=alpha$, $lambda_3=beta$.
answered 8 hours ago
MindlackMindlack
5,413413
$endgroup$
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
1
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Note that the assumptions imply that if $L(x)=0$, then $x=0$. Besides, $|L(x)|=|x|$ for all $x$.
Now, note that $|a+b|=sqrt^2+$.
As a consequence, for all $x,y$, $langle L(x),, L(y) rangle=langle x,y rangle$.
Therefore, for any scalars $lambda_i$, for any vectors $x_i$, $$|sum_ilambda_iL(x_i)|=|sum_ilambda_ix_i|.$$
Then take $x_1=alpha u+beta v$, $x_2=u$, $x_3=v$, $lambda_1=-1$, $lambda_2=alpha$, $lambda_3=beta$.
answered 8 hours ago
MindlackMindlack
5,413413
$endgroup$
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
1
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
add a comment |
$begingroup$
Note that the assumptions imply that if $L(x)=0$, then $x=0$. Besides, $|L(x)|=|x|$ for all $x$.
Now, note that $|a+b|=sqrt^2+$.
As a consequence, for all $x,y$, $langle L(x),, L(y) rangle=langle x,y rangle$.
Therefore, for any scalars $lambda_i$, for any vectors $x_i$, $$|sum_ilambda_iL(x_i)|=|sum_ilambda_ix_i|.$$
Then take $x_1=alpha u+beta v$, $x_2=u$, $x_3=v$, $lambda_1=-1$, $lambda_2=alpha$, $lambda_3=beta$.
answered 8 hours ago
MindlackMindlack
5,413413
$endgroup$
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
1
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
add a comment |
$begingroup$
Note that the assumptions imply that if $L(x)=0$, then $x=0$. Besides, $|L(x)|=|x|$ for all $x$.
Now, note that $|a+b|=sqrt^2+$.
As a consequence, for all $x,y$, $langle L(x),, L(y) rangle=langle x,y rangle$.
Therefore, for any scalars $lambda_i$, for any vectors $x_i$, $$|sum_ilambda_iL(x_i)|=|sum_ilambda_ix_i|.$$
Then take $x_1=alpha u+beta v$, $x_2=u$, $x_3=v$, $lambda_1=-1$, $lambda_2=alpha$, $lambda_3=beta$.
answered 8 hours ago
MindlackMindlack
5,413413
$endgroup$
Note that the assumptions imply that if $L(x)=0$, then $x=0$. Besides, $|L(x)|=|x|$ for all $x$.
Now, note that $|a+b|=sqrt^2+$.
As a consequence, for all $x,y$, $langle L(x),, L(y) rangle=langle x,y rangle$.
Therefore, for any scalars $lambda_i$, for any vectors $x_i$, $$|sum_ilambda_iL(x_i)|=|sum_ilambda_ix_i|.$$
Then take $x_1=alpha u+beta v$, $x_2=u$, $x_3=v$, $lambda_1=-1$, $lambda_2=alpha$, $lambda_3=beta$.
answered 8 hours ago
MindlackMindlack
5,413413
answered 8 hours ago
MindlackMindlack
5,413413
answered 8 hours ago
MindlackMindlack
5,413413
answered 8 hours ago
MindlackMindlack
5,413413
5,413413
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
1
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
add a comment |
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
1
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
$begingroup$
Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones.
$endgroup$
– EBP
7 hours ago
1
1
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
$begingroup$
Nevermind! They are equivalent. Thank you very much!
$endgroup$
– EBP
7 hours ago
add a comment |
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1
$begingroup$
Is the scalar field $mathbbR$ or $mathbbC$?
$endgroup$
– user159517
8 hours ago
$begingroup$
Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks.
$endgroup$
– EBP
8 hours ago
$begingroup$
Are we also assuming that $L$ is surjective?
$endgroup$
– user159517
8 hours ago
$begingroup$
No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity.
$endgroup$
– EBP
8 hours ago