Scalar Products and Projections The Next CEO of Stack OverflowCan a vector subspace have a unique complement in absence of choice?Representation theoremBilinear forms and scalar productScalars determine a vector in inner product space.Proof of uniqueness of identity element of addition of vector spaceAnother proof of uniqueness of identity element of addition of vector spaceOn dual spaces and inner productsDistance and uniqueness for projectionsScalar extension via tensor productRiesz Representation Theorem and Uniqueness
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Scalar Products and Projections
The Next CEO of Stack OverflowCan a vector subspace have a unique complement in absence of choice?Representation theoremBilinear forms and scalar productScalars determine a vector in inner product space.Proof of uniqueness of identity element of addition of vector spaceAnother proof of uniqueness of identity element of addition of vector spaceOn dual spaces and inner productsDistance and uniqueness for projectionsScalar extension via tensor productRiesz Representation Theorem and Uniqueness
$begingroup$
Proposition: Let w $in V$ so that V is a vector space over $mathbbR$ and $||w|| neq 0$. For every v in V there exists a unique $cin mathbbR$ so that $v-cw$ is perpendicular to w.
My Proof:
Existence: Assume $w in V$ and $||w|| neq 0$. Let $v in V$ be arbitrary. Seting c $in mathbbR$ to be $frac<v,w><w,w>$ we have $<v-cw,w>=<v-frac<v,w>w<w,w>,w>$ $=$ $<v,w>$ $-frac<v,w><w,w>$ $<w,w>$ $=$ $<v,w>(1-1)=0$
Uniqueness: Assume $<v-cw,w>=0$ $=$ $<v,w>-<cw,w>$ $=$ $<v,w>-c<w,w>=0$ $implies$ $c=frac<v,w><w,w>$.
Is the proof correct? How do I improve it? In particular i'm concerned with the uniqueness part.
linear-algebra vector-spaces vector-analysis inner-product-space orthogonality
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
$endgroup$
add a comment |
$begingroup$
Proposition: Let w $in V$ so that V is a vector space over $mathbbR$ and $||w|| neq 0$. For every v in V there exists a unique $cin mathbbR$ so that $v-cw$ is perpendicular to w.
My Proof:
Existence: Assume $w in V$ and $||w|| neq 0$. Let $v in V$ be arbitrary. Seting c $in mathbbR$ to be $frac<v,w><w,w>$ we have $<v-cw,w>=<v-frac<v,w>w<w,w>,w>$ $=$ $<v,w>$ $-frac<v,w><w,w>$ $<w,w>$ $=$ $<v,w>(1-1)=0$
Uniqueness: Assume $<v-cw,w>=0$ $=$ $<v,w>-<cw,w>$ $=$ $<v,w>-c<w,w>=0$ $implies$ $c=frac<v,w><w,w>$.
Is the proof correct? How do I improve it? In particular i'm concerned with the uniqueness part.
linear-algebra vector-spaces vector-analysis inner-product-space orthogonality
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
$endgroup$
1
$begingroup$
I don't see a problem with the proof. Have you considered projection theorem since $ Bbb R$ is Banach and V convex?
$endgroup$
– PerelMan
Mar 27 at 19:07
add a comment |
$begingroup$
Proposition: Let w $in V$ so that V is a vector space over $mathbbR$ and $||w|| neq 0$. For every v in V there exists a unique $cin mathbbR$ so that $v-cw$ is perpendicular to w.
My Proof:
Existence: Assume $w in V$ and $||w|| neq 0$. Let $v in V$ be arbitrary. Seting c $in mathbbR$ to be $frac<v,w><w,w>$ we have $<v-cw,w>=<v-frac<v,w>w<w,w>,w>$ $=$ $<v,w>$ $-frac<v,w><w,w>$ $<w,w>$ $=$ $<v,w>(1-1)=0$
Uniqueness: Assume $<v-cw,w>=0$ $=$ $<v,w>-<cw,w>$ $=$ $<v,w>-c<w,w>=0$ $implies$ $c=frac<v,w><w,w>$.
Is the proof correct? How do I improve it? In particular i'm concerned with the uniqueness part.
linear-algebra vector-spaces vector-analysis inner-product-space orthogonality
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
$endgroup$
Proposition: Let w $in V$ so that V is a vector space over $mathbbR$ and $||w|| neq 0$. For every v in V there exists a unique $cin mathbbR$ so that $v-cw$ is perpendicular to w.
My Proof:
Existence: Assume $w in V$ and $||w|| neq 0$. Let $v in V$ be arbitrary. Seting c $in mathbbR$ to be $frac<v,w><w,w>$ we have $<v-cw,w>=<v-frac<v,w>w<w,w>,w>$ $=$ $<v,w>$ $-frac<v,w><w,w>$ $<w,w>$ $=$ $<v,w>(1-1)=0$
Uniqueness: Assume $<v-cw,w>=0$ $=$ $<v,w>-<cw,w>$ $=$ $<v,w>-c<w,w>=0$ $implies$ $c=frac<v,w><w,w>$.
Is the proof correct? How do I improve it? In particular i'm concerned with the uniqueness part.
linear-algebra vector-spaces vector-analysis inner-product-space orthogonality
linear-algebra vector-spaces vector-analysis inner-product-space orthogonality
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
asked Mar 27 at 18:58
topologicalmagiciantopologicalmagician
1199
1199
1
$begingroup$
I don't see a problem with the proof. Have you considered projection theorem since $ Bbb R$ is Banach and V convex?
$endgroup$
– PerelMan
Mar 27 at 19:07
add a comment |
1
$begingroup$
I don't see a problem with the proof. Have you considered projection theorem since $ Bbb R$ is Banach and V convex?
$endgroup$
– PerelMan
Mar 27 at 19:07
1
1
$begingroup$
I don't see a problem with the proof. Have you considered projection theorem since $ Bbb R$ is Banach and V convex?
$endgroup$
– PerelMan
Mar 27 at 19:07
$begingroup$
I don't see a problem with the proof. Have you considered projection theorem since $ Bbb R$ is Banach and V convex?
$endgroup$
– PerelMan
Mar 27 at 19:07
add a comment |
1 Answer
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$begingroup$
In is not particulary well written, but it is correct. In the existence proof, after the third $=$ sign I would put $langle v,wrangle-langle v,wrangle=0$; it looks more natural to me.
In the uniqueness part, I would begin with “Assume that $langle v-cw,wrangle=0$. Then…”
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
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$begingroup$
In is not particulary well written, but it is correct. In the existence proof, after the third $=$ sign I would put $langle v,wrangle-langle v,wrangle=0$; it looks more natural to me.
In the uniqueness part, I would begin with “Assume that $langle v-cw,wrangle=0$. Then…”
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
In is not particulary well written, but it is correct. In the existence proof, after the third $=$ sign I would put $langle v,wrangle-langle v,wrangle=0$; it looks more natural to me.
In the uniqueness part, I would begin with “Assume that $langle v-cw,wrangle=0$. Then…”
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
In is not particulary well written, but it is correct. In the existence proof, after the third $=$ sign I would put $langle v,wrangle-langle v,wrangle=0$; it looks more natural to me.
In the uniqueness part, I would begin with “Assume that $langle v-cw,wrangle=0$. Then…”
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
In is not particulary well written, but it is correct. In the existence proof, after the third $=$ sign I would put $langle v,wrangle-langle v,wrangle=0$; it looks more natural to me.
In the uniqueness part, I would begin with “Assume that $langle v-cw,wrangle=0$. Then…”
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
edited Mar 27 at 19:04
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 27 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
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I don't see a problem with the proof. Have you considered projection theorem since $ Bbb R$ is Banach and V convex?
$endgroup$
– PerelMan
Mar 27 at 19:07