Vector space is decomposed into direct sum of its subspace and its orthogonal complement
$begingroup$
Let $E$ be a finite-dimensional vector space over the field $k$ and let $g$ be a bilinear form which is symmetric, antisymmetric or hermitian. Let $V$ be a subspace of $E$. Let $V_1 = { x| x in E, g(x,a)=0 forall a in V}$. Prove that $E$ is a direct sum of $V$ and $V_1$. This result is used in the proof as well-known but unfortunately I can't find the proof of it. The proof I know is in the case of $g$ being inner product.
vector-spaces bilinear-form
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
$endgroup$
add a comment |
$begingroup$
Let $E$ be a finite-dimensional vector space over the field $k$ and let $g$ be a bilinear form which is symmetric, antisymmetric or hermitian. Let $V$ be a subspace of $E$. Let $V_1 = { x| x in E, g(x,a)=0 forall a in V}$. Prove that $E$ is a direct sum of $V$ and $V_1$. This result is used in the proof as well-known but unfortunately I can't find the proof of it. The proof I know is in the case of $g$ being inner product.
vector-spaces bilinear-form
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
$endgroup$
2
$begingroup$
Well, if you can't "find the proof" come up with your own! Or at least try...
$endgroup$
– DonAntonio
Nov 7 '13 at 20:29
1
$begingroup$
This is false as stated, you need some kind of non-deneracy hypothesis on $g$. For example, if $g(x,y) = 0$ for all $x$ and $y$, then $g$ is symmetric. Now, let $V = E$. Then $V_1 = E$, but $Enotcong Eoplus E$ (for finite dimensional vector spaces).
$endgroup$
– Jason DeVito
Oct 14 '15 at 14:18
add a comment |
$begingroup$
Let $E$ be a finite-dimensional vector space over the field $k$ and let $g$ be a bilinear form which is symmetric, antisymmetric or hermitian. Let $V$ be a subspace of $E$. Let $V_1 = { x| x in E, g(x,a)=0 forall a in V}$. Prove that $E$ is a direct sum of $V$ and $V_1$. This result is used in the proof as well-known but unfortunately I can't find the proof of it. The proof I know is in the case of $g$ being inner product.
vector-spaces bilinear-form
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
$endgroup$
Let $E$ be a finite-dimensional vector space over the field $k$ and let $g$ be a bilinear form which is symmetric, antisymmetric or hermitian. Let $V$ be a subspace of $E$. Let $V_1 = { x| x in E, g(x,a)=0 forall a in V}$. Prove that $E$ is a direct sum of $V$ and $V_1$. This result is used in the proof as well-known but unfortunately I can't find the proof of it. The proof I know is in the case of $g$ being inner product.
vector-spaces bilinear-form
vector-spaces bilinear-form
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
edited Dec 11 '16 at 3:01
Gerry Myerson
147k8149302
147k8149302
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
asked Nov 7 '13 at 20:28
Nazar SerdyukNazar Serdyuk
1247
1247
2
$begingroup$
Well, if you can't "find the proof" come up with your own! Or at least try...
$endgroup$
– DonAntonio
Nov 7 '13 at 20:29
1
$begingroup$
This is false as stated, you need some kind of non-deneracy hypothesis on $g$. For example, if $g(x,y) = 0$ for all $x$ and $y$, then $g$ is symmetric. Now, let $V = E$. Then $V_1 = E$, but $Enotcong Eoplus E$ (for finite dimensional vector spaces).
$endgroup$
– Jason DeVito
Oct 14 '15 at 14:18
add a comment |
2
$begingroup$
Well, if you can't "find the proof" come up with your own! Or at least try...
$endgroup$
– DonAntonio
Nov 7 '13 at 20:29
1
$begingroup$
This is false as stated, you need some kind of non-deneracy hypothesis on $g$. For example, if $g(x,y) = 0$ for all $x$ and $y$, then $g$ is symmetric. Now, let $V = E$. Then $V_1 = E$, but $Enotcong Eoplus E$ (for finite dimensional vector spaces).
$endgroup$
– Jason DeVito
Oct 14 '15 at 14:18
2
2
$begingroup$
Well, if you can't "find the proof" come up with your own! Or at least try...
$endgroup$
– DonAntonio
Nov 7 '13 at 20:29
$begingroup$
Well, if you can't "find the proof" come up with your own! Or at least try...
$endgroup$
– DonAntonio
Nov 7 '13 at 20:29
1
1
$begingroup$
This is false as stated, you need some kind of non-deneracy hypothesis on $g$. For example, if $g(x,y) = 0$ for all $x$ and $y$, then $g$ is symmetric. Now, let $V = E$. Then $V_1 = E$, but $Enotcong Eoplus E$ (for finite dimensional vector spaces).
$endgroup$
– Jason DeVito
Oct 14 '15 at 14:18
$begingroup$
This is false as stated, you need some kind of non-deneracy hypothesis on $g$. For example, if $g(x,y) = 0$ for all $x$ and $y$, then $g$ is symmetric. Now, let $V = E$. Then $V_1 = E$, but $Enotcong Eoplus E$ (for finite dimensional vector spaces).
$endgroup$
– Jason DeVito
Oct 14 '15 at 14:18
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Hint in the finite dimensional case. Let $;n=dim_kE=n;$ and take a basis of $;V_1;,;{v_1,..,v_k}$ . Complete this to a basis of the whole space: $;{v_1,..,v_k,v_{k+1},...,v_n};$ . Using Gram-Schmidt we can assume this is an orthonormal basis, and then
$${v_{k+1},...,v_n};;text{is a basis of};;V$$
In fact, I can't see how this is really different from the usual proof for inner products
...
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
add a comment |
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$begingroup$
Hint in the finite dimensional case. Let $;n=dim_kE=n;$ and take a basis of $;V_1;,;{v_1,..,v_k}$ . Complete this to a basis of the whole space: $;{v_1,..,v_k,v_{k+1},...,v_n};$ . Using Gram-Schmidt we can assume this is an orthonormal basis, and then
$${v_{k+1},...,v_n};;text{is a basis of};;V$$
In fact, I can't see how this is really different from the usual proof for inner products
...
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
add a comment |
$begingroup$
Hint in the finite dimensional case. Let $;n=dim_kE=n;$ and take a basis of $;V_1;,;{v_1,..,v_k}$ . Complete this to a basis of the whole space: $;{v_1,..,v_k,v_{k+1},...,v_n};$ . Using Gram-Schmidt we can assume this is an orthonormal basis, and then
$${v_{k+1},...,v_n};;text{is a basis of};;V$$
In fact, I can't see how this is really different from the usual proof for inner products
...
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
add a comment |
$begingroup$
Hint in the finite dimensional case. Let $;n=dim_kE=n;$ and take a basis of $;V_1;,;{v_1,..,v_k}$ . Complete this to a basis of the whole space: $;{v_1,..,v_k,v_{k+1},...,v_n};$ . Using Gram-Schmidt we can assume this is an orthonormal basis, and then
$${v_{k+1},...,v_n};;text{is a basis of};;V$$
In fact, I can't see how this is really different from the usual proof for inner products
...
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
$endgroup$
Hint in the finite dimensional case. Let $;n=dim_kE=n;$ and take a basis of $;V_1;,;{v_1,..,v_k}$ . Complete this to a basis of the whole space: $;{v_1,..,v_k,v_{k+1},...,v_n};$ . Using Gram-Schmidt we can assume this is an orthonormal basis, and then
$${v_{k+1},...,v_n};;text{is a basis of};;V$$
In fact, I can't see how this is really different from the usual proof for inner products
...
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
edited Dec 30 '17 at 22:21
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
answered Nov 7 '13 at 20:34
DonAntonioDonAntonio
179k1494233
179k1494233
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
add a comment |
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
Don't we need to follow a Gram-Schmidt-like process to have ${v_{k+1},ldots,v_n}subset V$?
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:37
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
No...why would we?
$endgroup$
– DonAntonio
Nov 7 '13 at 20:38
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
Well, if $Vsubsetmathbb{R}^2$ is $Span{(1,0)}$, and we take the standard inner product, then ${(0,1)}$ is a basis for $V_1$. One could complete it to a basis for $mathbb{R}^2$ in the following way: ${(0,1),(1,1)}$, but naturally $(1,1)notin V$.
$endgroup$
– Jonathan Y.
Nov 7 '13 at 20:44
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
$begingroup$
I think you have a point there, @JonathanY. I shall edit later, perhaps...or delete as it is late now. Thanks.
$endgroup$
– DonAntonio
Nov 7 '13 at 20:48
add a comment |
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$begingroup$
Well, if you can't "find the proof" come up with your own! Or at least try...
$endgroup$
– DonAntonio
Nov 7 '13 at 20:29
1
$begingroup$
This is false as stated, you need some kind of non-deneracy hypothesis on $g$. For example, if $g(x,y) = 0$ for all $x$ and $y$, then $g$ is symmetric. Now, let $V = E$. Then $V_1 = E$, but $Enotcong Eoplus E$ (for finite dimensional vector spaces).
$endgroup$
– Jason DeVito
Oct 14 '15 at 14:18