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Prove vector space is the direct sum of subspace and its orthogonal complement

2

1

$begingroup$

$V$ is finite-dimensional over $BbbC$ and the form $langle cdot , cdot rangle$ is Hermitian. $U$ is a subspace of $V$.

Show that $V = U oplus U^perp$

I've been able to show that $U cap U^perp = 0$. I don't know how to approach showing that every vector $vin V$ can be written as $v = u + u'$, where $u, u'$ are in $U$ and $U^perp$ respectively. Any help would be much appreciated.

linear-algebra vector-spaces inner-product-space bilinear-form direct-sum

share|cite|improve this question

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

$endgroup$

add a comment | 

2

1

$begingroup$

$V$ is finite-dimensional over $BbbC$ and the form $langle cdot , cdot rangle$ is Hermitian. $U$ is a subspace of $V$.

Show that $V = U oplus U^perp$

I've been able to show that $U cap U^perp = 0$. I don't know how to approach showing that every vector $vin V$ can be written as $v = u + u'$, where $u, u'$ are in $U$ and $U^perp$ respectively. Any help would be much appreciated.

linear-algebra vector-spaces inner-product-space bilinear-form direct-sum

share|cite|improve this question

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

$endgroup$

add a comment | 

$begingroup$

$V$ is finite-dimensional over $BbbC$ and the form $langle cdot , cdot rangle$ is Hermitian. $U$ is a subspace of $V$.

Show that $V = U oplus U^perp$

I've been able to show that $U cap U^perp = 0$. I don't know how to approach showing that every vector $vin V$ can be written as $v = u + u'$, where $u, u'$ are in $U$ and $U^perp$ respectively. Any help would be much appreciated.

linear-algebra vector-spaces inner-product-space bilinear-form direct-sum

share|cite|improve this question

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

$endgroup$

share|cite|improve this question

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

share|cite|improve this question

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

edited Mar 27 at 13:40

José Carlos Santos

176k24134243

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

asked Dec 30 '17 at 22:12

mathsishardmathsishard

1113

1 Answer
1

active

oldest

votes

7

$begingroup$

Let $e_1,ldots,e_k$ be an orthonormal basis of $U$. For each $vin V$, let$$P(v)=sum_j=1^klangle v,e_jrangle e_j.$$Then$$(forall vin V):v=overbraceP(v)^in U+overbracebigl(v-P(v)bigr)^in U^perp.$$The fact that $v-P(v)in U^perp$ can be justified as follows: if $jin1,2,ldots,k$,beginalignbigllangle v-P(v),e_jbigrrangle=&leftlangle v-sum_l=1^klangle v,e_lrangle e_l,e_jrightrangle\&=langle v,e_jrangle-langle v,e_jrangle\&=0.endalignSince $e_1,ldots,e_k$ is a basis of $U$, this proves that $v-P(v)in U^perp$.

share|cite|improve this answer

edited Dec 30 '17 at 22:53

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243

$endgroup$

add a comment | 

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7

$begingroup$

Let $e_1,ldots,e_k$ be an orthonormal basis of $U$. For each $vin V$, let$$P(v)=sum_j=1^klangle v,e_jrangle e_j.$$Then$$(forall vin V):v=overbraceP(v)^in U+overbracebigl(v-P(v)bigr)^in U^perp.$$The fact that $v-P(v)in U^perp$ can be justified as follows: if $jin1,2,ldots,k$,beginalignbigllangle v-P(v),e_jbigrrangle=&leftlangle v-sum_l=1^klangle v,e_lrangle e_l,e_jrightrangle\&=langle v,e_jrangle-langle v,e_jrangle\&=0.endalignSince $e_1,ldots,e_k$ is a basis of $U$, this proves that $v-P(v)in U^perp$.

share|cite|improve this answer

edited Dec 30 '17 at 22:53

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243

$endgroup$

add a comment | 

7

$begingroup$

Let $e_1,ldots,e_k$ be an orthonormal basis of $U$. For each $vin V$, let$$P(v)=sum_j=1^klangle v,e_jrangle e_j.$$Then$$(forall vin V):v=overbraceP(v)^in U+overbracebigl(v-P(v)bigr)^in U^perp.$$The fact that $v-P(v)in U^perp$ can be justified as follows: if $jin1,2,ldots,k$,beginalignbigllangle v-P(v),e_jbigrrangle=&leftlangle v-sum_l=1^klangle v,e_lrangle e_l,e_jrightrangle\&=langle v,e_jrangle-langle v,e_jrangle\&=0.endalignSince $e_1,ldots,e_k$ is a basis of $U$, this proves that $v-P(v)in U^perp$.

share|cite|improve this answer

edited Dec 30 '17 at 22:53

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243

$endgroup$

add a comment | 

$begingroup$

Let $e_1,ldots,e_k$ be an orthonormal basis of $U$. For each $vin V$, let$$P(v)=sum_j=1^klangle v,e_jrangle e_j.$$Then$$(forall vin V):v=overbraceP(v)^in U+overbracebigl(v-P(v)bigr)^in U^perp.$$The fact that $v-P(v)in U^perp$ can be justified as follows: if $jin1,2,ldots,k$,beginalignbigllangle v-P(v),e_jbigrrangle=&leftlangle v-sum_l=1^klangle v,e_lrangle e_l,e_jrightrangle\&=langle v,e_jrangle-langle v,e_jrangle\&=0.endalignSince $e_1,ldots,e_k$ is a basis of $U$, this proves that $v-P(v)in U^perp$.

share|cite|improve this answer

edited Dec 30 '17 at 22:53

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243

$endgroup$

share|cite|improve this answer

edited Dec 30 '17 at 22:53

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243

answered Dec 30 '17 at 22:21

José Carlos SantosJosé Carlos Santos

176k24134243