Tricky Orthogonal Complement Lemma? 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)Orthogonal ComplementFinding the Orthogonal Complement to a subspaceWhat is the orthogonal complementHow to express double orthogonal complement?Prove $langle y,x rangle langle x,y rangle leq langle y,y rangle.$Proof: Sum of dimension of orthogonal complement and vector subspaceOrthogonal complementOrthogonal Complement of SpanDouble orthogonal complement of a finite dimensional subspaceFind orthogonal complement and its basis
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Tricky Orthogonal Complement Lemma?
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)Orthogonal ComplementFinding the Orthogonal Complement to a subspaceWhat is the orthogonal complementHow to express double orthogonal complement?Prove $langle y,x rangle langle x,y rangle leq langle y,y rangle.$Proof: Sum of dimension of orthogonal complement and vector subspaceOrthogonal complementOrthogonal Complement of SpanDouble orthogonal complement of a finite dimensional subspaceFind orthogonal complement and its basis
$begingroup$
I have the following definition:
Let $V$ be an inner product space and let $S subseteq V$.
The orthogonal complement of $S$, denoted $S^perp$, is
the set $$S^perp = leftvecv in V mid langlevecv, vecxrangle = 0 hspace1mm forallvecx in Sright.$$
It seems that it may be possible that if
$S^perp = leftvec0right$ then $textspan hspace0.2mm S = V$, but I'm having trouble proving if this is true or not. I've been trying to use the fact that $S^perp = left(textspan hspace0.2mm S right)^perp$.
linear-algebra
$endgroup$
add a comment |
$begingroup$
I have the following definition:
Let $V$ be an inner product space and let $S subseteq V$.
The orthogonal complement of $S$, denoted $S^perp$, is
the set $$S^perp = leftvecv in V mid langlevecv, vecxrangle = 0 hspace1mm forallvecx in Sright.$$
It seems that it may be possible that if
$S^perp = leftvec0right$ then $textspan hspace0.2mm S = V$, but I'm having trouble proving if this is true or not. I've been trying to use the fact that $S^perp = left(textspan hspace0.2mm S right)^perp$.
linear-algebra
$endgroup$
add a comment |
$begingroup$
I have the following definition:
Let $V$ be an inner product space and let $S subseteq V$.
The orthogonal complement of $S$, denoted $S^perp$, is
the set $$S^perp = leftvecv in V mid langlevecv, vecxrangle = 0 hspace1mm forallvecx in Sright.$$
It seems that it may be possible that if
$S^perp = leftvec0right$ then $textspan hspace0.2mm S = V$, but I'm having trouble proving if this is true or not. I've been trying to use the fact that $S^perp = left(textspan hspace0.2mm S right)^perp$.
linear-algebra
$endgroup$
I have the following definition:
Let $V$ be an inner product space and let $S subseteq V$.
The orthogonal complement of $S$, denoted $S^perp$, is
the set $$S^perp = leftvecv in V mid langlevecv, vecxrangle = 0 hspace1mm forallvecx in Sright.$$
It seems that it may be possible that if
$S^perp = leftvec0right$ then $textspan hspace0.2mm S = V$, but I'm having trouble proving if this is true or not. I've been trying to use the fact that $S^perp = left(textspan hspace0.2mm S right)^perp$.
linear-algebra
linear-algebra
asked Mar 31 at 6:49
Noam TobinNoam Tobin
82
82
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1 Answer
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$begingroup$
In general, we know that $(S^perp)^perp subseteq S$, where we always have equality in the finite-dimensional setting.
If $S^perp = 0$, then what is $(0)^perp$? What vectors satisfy $langle v, 0rangle = 0$?
Note that $(S^perp)^perp = (0)^perp subseteq Ssubseteq V$, and so $S=dots$
$endgroup$
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$begingroup$
In general, we know that $(S^perp)^perp subseteq S$, where we always have equality in the finite-dimensional setting.
If $S^perp = 0$, then what is $(0)^perp$? What vectors satisfy $langle v, 0rangle = 0$?
Note that $(S^perp)^perp = (0)^perp subseteq Ssubseteq V$, and so $S=dots$
$endgroup$
add a comment |
$begingroup$
In general, we know that $(S^perp)^perp subseteq S$, where we always have equality in the finite-dimensional setting.
If $S^perp = 0$, then what is $(0)^perp$? What vectors satisfy $langle v, 0rangle = 0$?
Note that $(S^perp)^perp = (0)^perp subseteq Ssubseteq V$, and so $S=dots$
$endgroup$
add a comment |
$begingroup$
In general, we know that $(S^perp)^perp subseteq S$, where we always have equality in the finite-dimensional setting.
If $S^perp = 0$, then what is $(0)^perp$? What vectors satisfy $langle v, 0rangle = 0$?
Note that $(S^perp)^perp = (0)^perp subseteq Ssubseteq V$, and so $S=dots$
$endgroup$
In general, we know that $(S^perp)^perp subseteq S$, where we always have equality in the finite-dimensional setting.
If $S^perp = 0$, then what is $(0)^perp$? What vectors satisfy $langle v, 0rangle = 0$?
Note that $(S^perp)^perp = (0)^perp subseteq Ssubseteq V$, and so $S=dots$
answered Mar 31 at 10:43
Santana AftonSantana Afton
3,0992730
3,0992730
add a comment |
add a comment |
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