Vector space is decomposed into direct sum of its subspace and its orthogonal complement












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '16 at 3:01









Gerry Myerson

147k8149302




147k8149302










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
















  • 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












1 Answer
1






active

oldest

votes


















0












$begingroup$

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
...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

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
...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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
...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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
...






share|cite|improve this answer











$endgroup$



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
...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '17 at 22:21

























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


















  • $begingroup$
    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


















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