Extending inner product from subspace to whole space

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Suppose that V is a vector space over either R or C, W is a
subspace of V , and we are given an inner product on W. Show that there
is at least one way to extend this function to an inner product on all of
V . Do not assume that V is finite-dimensional.
My best guess is saying V=W+U f(u+w)=w where f linear but that gives then $langle f(u+w),f(u+w) rangle$=0 if v is an element u or or v=0 which contradicts one of the axioms of an inner product
linear-algebra abstract-algebra inner-product-space
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up vote
-1
down vote
favorite
Suppose that V is a vector space over either R or C, W is a
subspace of V , and we are given an inner product on W. Show that there
is at least one way to extend this function to an inner product on all of
V . Do not assume that V is finite-dimensional.
My best guess is saying V=W+U f(u+w)=w where f linear but that gives then $langle f(u+w),f(u+w) rangle$=0 if v is an element u or or v=0 which contradicts one of the axioms of an inner product
linear-algebra abstract-algebra inner-product-space
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Suppose that V is a vector space over either R or C, W is a
subspace of V , and we are given an inner product on W. Show that there
is at least one way to extend this function to an inner product on all of
V . Do not assume that V is finite-dimensional.
My best guess is saying V=W+U f(u+w)=w where f linear but that gives then $langle f(u+w),f(u+w) rangle$=0 if v is an element u or or v=0 which contradicts one of the axioms of an inner product
linear-algebra abstract-algebra inner-product-space
Suppose that V is a vector space over either R or C, W is a
subspace of V , and we are given an inner product on W. Show that there
is at least one way to extend this function to an inner product on all of
V . Do not assume that V is finite-dimensional.
My best guess is saying V=W+U f(u+w)=w where f linear but that gives then $langle f(u+w),f(u+w) rangle$=0 if v is an element u or or v=0 which contradicts one of the axioms of an inner product
linear-algebra abstract-algebra inner-product-space
asked Aug 17 at 22:13
user3555068
43
43
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1 Answer
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Take a subspace $W'$ of $V$ such that $V=Wbigoplus W'$. Define an inner product $g$ on $W'$. If $f$ is the inner product on $W$, you can extend it to the whole space by$$langle v+v',w+w'rangle=f(v,w)+g(v',w').$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Take a subspace $W'$ of $V$ such that $V=Wbigoplus W'$. Define an inner product $g$ on $W'$. If $f$ is the inner product on $W$, you can extend it to the whole space by$$langle v+v',w+w'rangle=f(v,w)+g(v',w').$$
add a comment |Â
up vote
1
down vote
accepted
Take a subspace $W'$ of $V$ such that $V=Wbigoplus W'$. Define an inner product $g$ on $W'$. If $f$ is the inner product on $W$, you can extend it to the whole space by$$langle v+v',w+w'rangle=f(v,w)+g(v',w').$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Take a subspace $W'$ of $V$ such that $V=Wbigoplus W'$. Define an inner product $g$ on $W'$. If $f$ is the inner product on $W$, you can extend it to the whole space by$$langle v+v',w+w'rangle=f(v,w)+g(v',w').$$
Take a subspace $W'$ of $V$ such that $V=Wbigoplus W'$. Define an inner product $g$ on $W'$. If $f$ is the inner product on $W$, you can extend it to the whole space by$$langle v+v',w+w'rangle=f(v,w)+g(v',w').$$
edited Aug 17 at 22:29
answered Aug 17 at 22:23
José Carlos Santos
117k16101180
117k16101180
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