Mixing
Equality of dimensions of two space
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Let $T:H_1 to H_2$ be a bijective bounded linear operator between Hilbert spaces $H_1$ and $H_2$. Do the algebraic dimensions of $H_1$ and $H_2$ coincide? I think the answer is positive but I can not prove it. Please give me any regard. Also, please introduce any references about dimension of spaces.
linear-algebra functional-analysis
asked Jan 24 at 15:05
saeedsaeed
18410
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add a comment |
$begingroup$
Let $T:H_1 to H_2$ be a bijective bounded linear operator between Hilbert spaces $H_1$ and $H_2$. Do the algebraic dimensions of $H_1$ and $H_2$ coincide? I think the answer is positive but I can not prove it. Please give me any regard. Also, please introduce any references about dimension of spaces.
linear-algebra functional-analysis
asked Jan 24 at 15:05
saeedsaeed
18410
$endgroup$
add a comment |
$begingroup$
Let $T:H_1 to H_2$ be a bijective bounded linear operator between Hilbert spaces $H_1$ and $H_2$. Do the algebraic dimensions of $H_1$ and $H_2$ coincide? I think the answer is positive but I can not prove it. Please give me any regard. Also, please introduce any references about dimension of spaces.
linear-algebra functional-analysis
asked Jan 24 at 15:05
saeedsaeed
18410
$endgroup$
Let $T:H_1 to H_2$ be a bijective bounded linear operator between Hilbert spaces $H_1$ and $H_2$. Do the algebraic dimensions of $H_1$ and $H_2$ coincide? I think the answer is positive but I can not prove it. Please give me any regard. Also, please introduce any references about dimension of spaces.
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Jan 24 at 15:05
saeedsaeed
18410
asked Jan 24 at 15:05
saeedsaeed
18410
asked Jan 24 at 15:05
saeedsaeed
18410
asked Jan 24 at 15:05
saeedsaeed
18410
asked Jan 24 at 15:05
saeedsaeed
18410
18410
add a comment |
add a comment |
1 Answer
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If $H_1,H_2$ are finite dimensional this is a well known result in linear algebra.
If they're infinite dimensional then you need to define those notions. The most "natural" way to do that is to ask whether there is a bijection between a basis of $H_1$ and a basis of $H_2$.
The answer is yes: Let ${e_i:iin I}$ be an algebraic basis for $H_1$ (which exists using the axiom of choice), we claim that ${T(e_i):iin I}$ is a basis of $H_2$.
Proof: let $h_2in H_2$ as $T$ is a bijection there exists $h_1in H$ such that $T(h_1)=h_2$. As ${e_i:iin I}$ an algebraic basis for $H_1$ there exists finitely many indices $e_{i_1},...,e_{i_n}$ such that $h_1=sum_{k=1}^n c_{i_k}e_{i_k}$ and therefore $T(h_1) = sum_{k=1}^n c_{i_k}T(e_{i_k})$ we conclude that ${T(e_i):iin I}$ is a generating set.
It is clearly minimal, as if we can remove $T(e_i)$ for some $i$ then $T(e_i)$ is a linear combination of some $T(e_{j_1}),...,T(e_{j_m})$ but then since $T$ is injective we have that $e_i$ a linear combination of $e_{j_1},...,e_{j_m}$ which contradicts the fact that ${e_i:iin I}$ is minimal.
We conclude that ${T(e_i):iin I}$ is a minimal generating set, hence a basis.
answered Jan 24 at 15:14
YankoYanko
7,8801830
$endgroup$
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $H_1,H_2$ are finite dimensional this is a well known result in linear algebra.
If they're infinite dimensional then you need to define those notions. The most "natural" way to do that is to ask whether there is a bijection between a basis of $H_1$ and a basis of $H_2$.
The answer is yes: Let ${e_i:iin I}$ be an algebraic basis for $H_1$ (which exists using the axiom of choice), we claim that ${T(e_i):iin I}$ is a basis of $H_2$.
Proof: let $h_2in H_2$ as $T$ is a bijection there exists $h_1in H$ such that $T(h_1)=h_2$. As ${e_i:iin I}$ an algebraic basis for $H_1$ there exists finitely many indices $e_{i_1},...,e_{i_n}$ such that $h_1=sum_{k=1}^n c_{i_k}e_{i_k}$ and therefore $T(h_1) = sum_{k=1}^n c_{i_k}T(e_{i_k})$ we conclude that ${T(e_i):iin I}$ is a generating set.
It is clearly minimal, as if we can remove $T(e_i)$ for some $i$ then $T(e_i)$ is a linear combination of some $T(e_{j_1}),...,T(e_{j_m})$ but then since $T$ is injective we have that $e_i$ a linear combination of $e_{j_1},...,e_{j_m}$ which contradicts the fact that ${e_i:iin I}$ is minimal.
We conclude that ${T(e_i):iin I}$ is a minimal generating set, hence a basis.
answered Jan 24 at 15:14
YankoYanko
7,8801830
$endgroup$
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
add a comment |
$begingroup$
If $H_1,H_2$ are finite dimensional this is a well known result in linear algebra.
If they're infinite dimensional then you need to define those notions. The most "natural" way to do that is to ask whether there is a bijection between a basis of $H_1$ and a basis of $H_2$.
The answer is yes: Let ${e_i:iin I}$ be an algebraic basis for $H_1$ (which exists using the axiom of choice), we claim that ${T(e_i):iin I}$ is a basis of $H_2$.
Proof: let $h_2in H_2$ as $T$ is a bijection there exists $h_1in H$ such that $T(h_1)=h_2$. As ${e_i:iin I}$ an algebraic basis for $H_1$ there exists finitely many indices $e_{i_1},...,e_{i_n}$ such that $h_1=sum_{k=1}^n c_{i_k}e_{i_k}$ and therefore $T(h_1) = sum_{k=1}^n c_{i_k}T(e_{i_k})$ we conclude that ${T(e_i):iin I}$ is a generating set.
It is clearly minimal, as if we can remove $T(e_i)$ for some $i$ then $T(e_i)$ is a linear combination of some $T(e_{j_1}),...,T(e_{j_m})$ but then since $T$ is injective we have that $e_i$ a linear combination of $e_{j_1},...,e_{j_m}$ which contradicts the fact that ${e_i:iin I}$ is minimal.
We conclude that ${T(e_i):iin I}$ is a minimal generating set, hence a basis.
answered Jan 24 at 15:14
YankoYanko
7,8801830
$endgroup$
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
add a comment |
$begingroup$
If $H_1,H_2$ are finite dimensional this is a well known result in linear algebra.
If they're infinite dimensional then you need to define those notions. The most "natural" way to do that is to ask whether there is a bijection between a basis of $H_1$ and a basis of $H_2$.
The answer is yes: Let ${e_i:iin I}$ be an algebraic basis for $H_1$ (which exists using the axiom of choice), we claim that ${T(e_i):iin I}$ is a basis of $H_2$.
Proof: let $h_2in H_2$ as $T$ is a bijection there exists $h_1in H$ such that $T(h_1)=h_2$. As ${e_i:iin I}$ an algebraic basis for $H_1$ there exists finitely many indices $e_{i_1},...,e_{i_n}$ such that $h_1=sum_{k=1}^n c_{i_k}e_{i_k}$ and therefore $T(h_1) = sum_{k=1}^n c_{i_k}T(e_{i_k})$ we conclude that ${T(e_i):iin I}$ is a generating set.
It is clearly minimal, as if we can remove $T(e_i)$ for some $i$ then $T(e_i)$ is a linear combination of some $T(e_{j_1}),...,T(e_{j_m})$ but then since $T$ is injective we have that $e_i$ a linear combination of $e_{j_1},...,e_{j_m}$ which contradicts the fact that ${e_i:iin I}$ is minimal.
We conclude that ${T(e_i):iin I}$ is a minimal generating set, hence a basis.
answered Jan 24 at 15:14
YankoYanko
7,8801830
$endgroup$
If $H_1,H_2$ are finite dimensional this is a well known result in linear algebra.
If they're infinite dimensional then you need to define those notions. The most "natural" way to do that is to ask whether there is a bijection between a basis of $H_1$ and a basis of $H_2$.
The answer is yes: Let ${e_i:iin I}$ be an algebraic basis for $H_1$ (which exists using the axiom of choice), we claim that ${T(e_i):iin I}$ is a basis of $H_2$.
Proof: let $h_2in H_2$ as $T$ is a bijection there exists $h_1in H$ such that $T(h_1)=h_2$. As ${e_i:iin I}$ an algebraic basis for $H_1$ there exists finitely many indices $e_{i_1},...,e_{i_n}$ such that $h_1=sum_{k=1}^n c_{i_k}e_{i_k}$ and therefore $T(h_1) = sum_{k=1}^n c_{i_k}T(e_{i_k})$ we conclude that ${T(e_i):iin I}$ is a generating set.
It is clearly minimal, as if we can remove $T(e_i)$ for some $i$ then $T(e_i)$ is a linear combination of some $T(e_{j_1}),...,T(e_{j_m})$ but then since $T$ is injective we have that $e_i$ a linear combination of $e_{j_1},...,e_{j_m}$ which contradicts the fact that ${e_i:iin I}$ is minimal.
We conclude that ${T(e_i):iin I}$ is a minimal generating set, hence a basis.
answered Jan 24 at 15:14
YankoYanko
7,8801830
answered Jan 24 at 15:14
YankoYanko
7,8801830
answered Jan 24 at 15:14
YankoYanko
7,8801830
answered Jan 24 at 15:14
YankoYanko
7,8801830
7,8801830
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
add a comment |
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
$begingroup$
This is also true if we define dimension as the cardinal of an orthonormal (or Schauder) basis, but it requires some work, since one has to see that the inverse of a bounded bijective map is also bounded.
$endgroup$
– mlainz
Jan 24 at 15:17
add a comment |
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