The orthogonal complement of the orthogonal complement from “Linear Algebra Done Right” The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Orthogonality, proof in Linear Algebra/Functional AnalysiShow that $V = U^perp bigoplus U$Linear Algebra question about orthogonal projection (Upper Linear Algebra)Linear Algebra Proof confirmationOrthogonal Projection and Orthogonal Complement Are Orthogonal To One AnotherOrthogonal complementDouble orthogonal complement of a finite dimensional subspaceOrthogonal Complement of a Orthonormal BasisIf we have an orthonormal basis of $mathbbR^n$ how we can describe every vector in $mathbbRnT$ using them?Gram-Schmidt Procedure from “Linear Algebra Done Right”
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The orthogonal complement of the orthogonal complement from “Linear Algebra Done Right”
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Orthogonality, proof in Linear Algebra/Functional AnalysiShow that $V = U^perp bigoplus U$Linear Algebra question about orthogonal projection (Upper Linear Algebra)Linear Algebra Proof confirmationOrthogonal Projection and Orthogonal Complement Are Orthogonal To One AnotherOrthogonal complementDouble orthogonal complement of a finite dimensional subspaceOrthogonal Complement of a Orthonormal BasisIf we have an orthonormal basis of $mathbbR^n$ how we can describe every vector in $mathbbRnT$ using them?Gram-Schmidt Procedure from “Linear Algebra Done Right”
$begingroup$
The following content is from "Linear Algebra Done Right" by Sheldon Axler
Corollary: Suppose $U$ is a finite-dimensional subspace of $V$. Then $$U = (U ^perp)^perp.$$
We need to prove the following: i) $U subseteq (U^perp)^perp$ and
ii) $(U^perp)^perp subseteq U$
Proof i): Supposer $u in U$. Then $langle u,v rangle = 0$ for $v in U^perp$. Since all vector $u$ is orthogonal to $v$. Then $u in U$. Hence $u in (U ^perp)^perp$
I do understand the proof for i) as stated in the above.
But I'm confused about the proof in ii)
ii) Let $u in (U ^perp)^perp$ From Proposition 2, we have that $V = U bigoplus U^perp$. and so $u$ can be written as $u=v+w$ where $v in U$ and $w in U^perp$. But $u - v = w$, so $u - v in U^perp$.
Now we already have that $u in (U ^perp)^perp$ and $v in (U ^perp)^perp$ and $u - v in (U ^perp)^perp$. Therfore, $u - v in U ^perp cap (U^perp)^perp$.
***Here is where I don't understand! Why we can let $u = v + w$, but since $u in (U ^perp)^perp$? and Why does $u - v in U ^perp cap (U^perp)^perp$?
linear-algebra abstract-algebra proof-verification orthogonality
edited Mar 24 at 18:51
darij grinberg
11.5k33168
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
$endgroup$
add a comment |
$begingroup$
The following content is from "Linear Algebra Done Right" by Sheldon Axler
Corollary: Suppose $U$ is a finite-dimensional subspace of $V$. Then $$U = (U ^perp)^perp.$$
We need to prove the following: i) $U subseteq (U^perp)^perp$ and
ii) $(U^perp)^perp subseteq U$
Proof i): Supposer $u in U$. Then $langle u,v rangle = 0$ for $v in U^perp$. Since all vector $u$ is orthogonal to $v$. Then $u in U$. Hence $u in (U ^perp)^perp$
I do understand the proof for i) as stated in the above.
But I'm confused about the proof in ii)
ii) Let $u in (U ^perp)^perp$ From Proposition 2, we have that $V = U bigoplus U^perp$. and so $u$ can be written as $u=v+w$ where $v in U$ and $w in U^perp$. But $u - v = w$, so $u - v in U^perp$.
Now we already have that $u in (U ^perp)^perp$ and $v in (U ^perp)^perp$ and $u - v in (U ^perp)^perp$. Therfore, $u - v in U ^perp cap (U^perp)^perp$.
***Here is where I don't understand! Why we can let $u = v + w$, but since $u in (U ^perp)^perp$? and Why does $u - v in U ^perp cap (U^perp)^perp$?
linear-algebra abstract-algebra proof-verification orthogonality
edited Mar 24 at 18:51
darij grinberg
11.5k33168
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
$endgroup$
$begingroup$
Since $u - v in U ^perp $ and $u - v in (U^perp)^perp$, it follows that $u - v in U^perp cap (U^perp)^perp$
$endgroup$
– J. W. Tanner
Mar 24 at 16:51
$begingroup$
Thanks! I got how $u - v in U^perp cap (U^perp)^perp$ work right now. But I don't get why we can let $u = v + w$, but since $u in (U^perp)^perp$
$endgroup$
– Algorisum
Mar 24 at 16:52
$begingroup$
Wait, how is $U oplus U^perp = V$ ? How is $U^perp$ even a subspace of $V$? Are we in an inner product space?
$endgroup$
– darij grinberg
Mar 24 at 18:53
$begingroup$
You should declare what $V$ is, and in particular what $U^perp$ is. I suspect you mean $V$ to be an inner product space over real or complex numbers, and $U^perp$ to be the orthogonal complement, but in general, $U^perp$ may either not make sense, or $(U^perp)^perp$ may be much bigger than $U$.
$endgroup$
– tomasz
Mar 24 at 18:56
add a comment |
$begingroup$
The following content is from "Linear Algebra Done Right" by Sheldon Axler
Corollary: Suppose $U$ is a finite-dimensional subspace of $V$. Then $$U = (U ^perp)^perp.$$
We need to prove the following: i) $U subseteq (U^perp)^perp$ and
ii) $(U^perp)^perp subseteq U$
Proof i): Supposer $u in U$. Then $langle u,v rangle = 0$ for $v in U^perp$. Since all vector $u$ is orthogonal to $v$. Then $u in U$. Hence $u in (U ^perp)^perp$
I do understand the proof for i) as stated in the above.
But I'm confused about the proof in ii)
ii) Let $u in (U ^perp)^perp$ From Proposition 2, we have that $V = U bigoplus U^perp$. and so $u$ can be written as $u=v+w$ where $v in U$ and $w in U^perp$. But $u - v = w$, so $u - v in U^perp$.
Now we already have that $u in (U ^perp)^perp$ and $v in (U ^perp)^perp$ and $u - v in (U ^perp)^perp$. Therfore, $u - v in U ^perp cap (U^perp)^perp$.
***Here is where I don't understand! Why we can let $u = v + w$, but since $u in (U ^perp)^perp$? and Why does $u - v in U ^perp cap (U^perp)^perp$?
linear-algebra abstract-algebra proof-verification orthogonality
edited Mar 24 at 18:51
darij grinberg
11.5k33168
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
$endgroup$
The following content is from "Linear Algebra Done Right" by Sheldon Axler
Corollary: Suppose $U$ is a finite-dimensional subspace of $V$. Then $$U = (U ^perp)^perp.$$
We need to prove the following: i) $U subseteq (U^perp)^perp$ and
ii) $(U^perp)^perp subseteq U$
Proof i): Supposer $u in U$. Then $langle u,v rangle = 0$ for $v in U^perp$. Since all vector $u$ is orthogonal to $v$. Then $u in U$. Hence $u in (U ^perp)^perp$
I do understand the proof for i) as stated in the above.
But I'm confused about the proof in ii)
ii) Let $u in (U ^perp)^perp$ From Proposition 2, we have that $V = U bigoplus U^perp$. and so $u$ can be written as $u=v+w$ where $v in U$ and $w in U^perp$. But $u - v = w$, so $u - v in U^perp$.
Now we already have that $u in (U ^perp)^perp$ and $v in (U ^perp)^perp$ and $u - v in (U ^perp)^perp$. Therfore, $u - v in U ^perp cap (U^perp)^perp$.
***Here is where I don't understand! Why we can let $u = v + w$, but since $u in (U ^perp)^perp$? and Why does $u - v in U ^perp cap (U^perp)^perp$?
linear-algebra abstract-algebra proof-verification orthogonality
linear-algebra abstract-algebra proof-verification orthogonality
edited Mar 24 at 18:51
darij grinberg
11.5k33168
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
edited Mar 24 at 18:51
darij grinberg
11.5k33168
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
edited Mar 24 at 18:51
darij grinberg
11.5k33168
edited Mar 24 at 18:51
darij grinberg
11.5k33168
edited Mar 24 at 18:51
darij grinberg
11.5k33168
11.5k33168
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
asked Mar 24 at 16:35
AlgorisumAlgorisum
457
457
$begingroup$
Since $u - v in U ^perp $ and $u - v in (U^perp)^perp$, it follows that $u - v in U^perp cap (U^perp)^perp$
$endgroup$
– J. W. Tanner
Mar 24 at 16:51
$begingroup$
Thanks! I got how $u - v in U^perp cap (U^perp)^perp$ work right now. But I don't get why we can let $u = v + w$, but since $u in (U^perp)^perp$
$endgroup$
– Algorisum
Mar 24 at 16:52
$begingroup$
Wait, how is $U oplus U^perp = V$ ? How is $U^perp$ even a subspace of $V$? Are we in an inner product space?
$endgroup$
– darij grinberg
Mar 24 at 18:53
$begingroup$
You should declare what $V$ is, and in particular what $U^perp$ is. I suspect you mean $V$ to be an inner product space over real or complex numbers, and $U^perp$ to be the orthogonal complement, but in general, $U^perp$ may either not make sense, or $(U^perp)^perp$ may be much bigger than $U$.
$endgroup$
– tomasz
Mar 24 at 18:56
add a comment |
$begingroup$
Since $u - v in U ^perp $ and $u - v in (U^perp)^perp$, it follows that $u - v in U^perp cap (U^perp)^perp$
$endgroup$
– J. W. Tanner
Mar 24 at 16:51
$begingroup$
Thanks! I got how $u - v in U^perp cap (U^perp)^perp$ work right now. But I don't get why we can let $u = v + w$, but since $u in (U^perp)^perp$
$endgroup$
– Algorisum
Mar 24 at 16:52
$begingroup$
Wait, how is $U oplus U^perp = V$ ? How is $U^perp$ even a subspace of $V$? Are we in an inner product space?
$endgroup$
– darij grinberg
Mar 24 at 18:53
$begingroup$
You should declare what $V$ is, and in particular what $U^perp$ is. I suspect you mean $V$ to be an inner product space over real or complex numbers, and $U^perp$ to be the orthogonal complement, but in general, $U^perp$ may either not make sense, or $(U^perp)^perp$ may be much bigger than $U$.
$endgroup$
– tomasz
Mar 24 at 18:56
$begingroup$
Since $u - v in U ^perp $ and $u - v in (U^perp)^perp$, it follows that $u - v in U^perp cap (U^perp)^perp$
$endgroup$
– J. W. Tanner
Mar 24 at 16:51
$begingroup$
Since $u - v in U ^perp $ and $u - v in (U^perp)^perp$, it follows that $u - v in U^perp cap (U^perp)^perp$
$endgroup$
– J. W. Tanner
Mar 24 at 16:51
$begingroup$
Thanks! I got how $u - v in U^perp cap (U^perp)^perp$ work right now. But I don't get why we can let $u = v + w$, but since $u in (U^perp)^perp$
$endgroup$
– Algorisum
Mar 24 at 16:52
$begingroup$
Thanks! I got how $u - v in U^perp cap (U^perp)^perp$ work right now. But I don't get why we can let $u = v + w$, but since $u in (U^perp)^perp$
$endgroup$
– Algorisum
Mar 24 at 16:52
$begingroup$
Wait, how is $U oplus U^perp = V$ ? How is $U^perp$ even a subspace of $V$? Are we in an inner product space?
$endgroup$
– darij grinberg
Mar 24 at 18:53
$begingroup$
Wait, how is $U oplus U^perp = V$ ? How is $U^perp$ even a subspace of $V$? Are we in an inner product space?
$endgroup$
– darij grinberg
Mar 24 at 18:53
$begingroup$
You should declare what $V$ is, and in particular what $U^perp$ is. I suspect you mean $V$ to be an inner product space over real or complex numbers, and $U^perp$ to be the orthogonal complement, but in general, $U^perp$ may either not make sense, or $(U^perp)^perp$ may be much bigger than $U$.
$endgroup$
– tomasz
Mar 24 at 18:56
$begingroup$
You should declare what $V$ is, and in particular what $U^perp$ is. I suspect you mean $V$ to be an inner product space over real or complex numbers, and $U^perp$ to be the orthogonal complement, but in general, $U^perp$ may either not make sense, or $(U^perp)^perp$ may be much bigger than $U$.
$endgroup$
– tomasz
Mar 24 at 18:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $V=Uoplus U^perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^perp$.
This holds, in particular, for $uin(U^perp)^perp$. So, according to Axler, write
$$
u=v+w,qquad vin U, win U^perp
$$
Now apply the assumption that $uin(U^perp)^perp$ and the fact that $vin Usubseteq(U^perp)^perp$ to conclude that
$$
w=u-vin(U^perp)^perp
$$
as well. On the other hand, $win U^perp$ by hypothesis. Therefore
$$
w=u-vin U^perpcap(U^perp)^perp
$$
A general result about orthogonal complements is that, for every subspace $X$, $Xcap X^perp=0$. This holds in particular for $X=U^perp$.
Hence $w=u-v=0$, so $u=vin U$.
answered Mar 24 at 16:53
egregegreg
186k1486208
$endgroup$
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
Since $V=Uoplus U^perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^perp$.
This holds, in particular, for $uin(U^perp)^perp$. So, according to Axler, write
$$
u=v+w,qquad vin U, win U^perp
$$
Now apply the assumption that $uin(U^perp)^perp$ and the fact that $vin Usubseteq(U^perp)^perp$ to conclude that
$$
w=u-vin(U^perp)^perp
$$
as well. On the other hand, $win U^perp$ by hypothesis. Therefore
$$
w=u-vin U^perpcap(U^perp)^perp
$$
A general result about orthogonal complements is that, for every subspace $X$, $Xcap X^perp=0$. This holds in particular for $X=U^perp$.
Hence $w=u-v=0$, so $u=vin U$.
answered Mar 24 at 16:53
egregegreg
186k1486208
$endgroup$
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
add a comment |
$begingroup$
Since $V=Uoplus U^perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^perp$.
This holds, in particular, for $uin(U^perp)^perp$. So, according to Axler, write
$$
u=v+w,qquad vin U, win U^perp
$$
Now apply the assumption that $uin(U^perp)^perp$ and the fact that $vin Usubseteq(U^perp)^perp$ to conclude that
$$
w=u-vin(U^perp)^perp
$$
as well. On the other hand, $win U^perp$ by hypothesis. Therefore
$$
w=u-vin U^perpcap(U^perp)^perp
$$
A general result about orthogonal complements is that, for every subspace $X$, $Xcap X^perp=0$. This holds in particular for $X=U^perp$.
Hence $w=u-v=0$, so $u=vin U$.
answered Mar 24 at 16:53
egregegreg
186k1486208
$endgroup$
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
add a comment |
$begingroup$
Since $V=Uoplus U^perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^perp$.
This holds, in particular, for $uin(U^perp)^perp$. So, according to Axler, write
$$
u=v+w,qquad vin U, win U^perp
$$
Now apply the assumption that $uin(U^perp)^perp$ and the fact that $vin Usubseteq(U^perp)^perp$ to conclude that
$$
w=u-vin(U^perp)^perp
$$
as well. On the other hand, $win U^perp$ by hypothesis. Therefore
$$
w=u-vin U^perpcap(U^perp)^perp
$$
A general result about orthogonal complements is that, for every subspace $X$, $Xcap X^perp=0$. This holds in particular for $X=U^perp$.
Hence $w=u-v=0$, so $u=vin U$.
answered Mar 24 at 16:53
egregegreg
186k1486208
$endgroup$
Since $V=Uoplus U^perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^perp$.
This holds, in particular, for $uin(U^perp)^perp$. So, according to Axler, write
$$
u=v+w,qquad vin U, win U^perp
$$
Now apply the assumption that $uin(U^perp)^perp$ and the fact that $vin Usubseteq(U^perp)^perp$ to conclude that
$$
w=u-vin(U^perp)^perp
$$
as well. On the other hand, $win U^perp$ by hypothesis. Therefore
$$
w=u-vin U^perpcap(U^perp)^perp
$$
A general result about orthogonal complements is that, for every subspace $X$, $Xcap X^perp=0$. This holds in particular for $X=U^perp$.
Hence $w=u-v=0$, so $u=vin U$.
answered Mar 24 at 16:53
egregegreg
186k1486208
answered Mar 24 at 16:53
egregegreg
186k1486208
answered Mar 24 at 16:53
egregegreg
186k1486208
answered Mar 24 at 16:53
egregegreg
186k1486208
186k1486208
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
add a comment |
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
Since $V = U oplus U^perp$, then U is a subspace of vector space V. But why does $u in (U^perp)^perp$ be equal to V, Since $u = v + w$?
$endgroup$
– Algorisum
Mar 24 at 16:59
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
@Algorisum $u$ is an element of $V$. Every element of $V$ can be written as $v+w$ for some $vin U$ and $win U^perp$.
$endgroup$
– egreg
Mar 24 at 17:03
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
$begingroup$
Yes.... I forget about that. Thank you
$endgroup$
– Algorisum
Mar 24 at 17:05
add a comment |
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Since $u - v in U ^perp $ and $u - v in (U^perp)^perp$, it follows that $u - v in U^perp cap (U^perp)^perp$
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– J. W. Tanner
Mar 24 at 16:51
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Thanks! I got how $u - v in U^perp cap (U^perp)^perp$ work right now. But I don't get why we can let $u = v + w$, but since $u in (U^perp)^perp$
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– Algorisum
Mar 24 at 16:52
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Wait, how is $U oplus U^perp = V$ ? How is $U^perp$ even a subspace of $V$? Are we in an inner product space?
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– darij grinberg
Mar 24 at 18:53
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You should declare what $V$ is, and in particular what $U^perp$ is. I suspect you mean $V$ to be an inner product space over real or complex numbers, and $U^perp$ to be the orthogonal complement, but in general, $U^perp$ may either not make sense, or $(U^perp)^perp$ may be much bigger than $U$.
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– tomasz
Mar 24 at 18:56