outer in terms of basis
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Let $v_1,v_2,v_3,...,v_n$ be an orthonormal basis of $V$. Show that for any vectors $w$ and $z$ of $V$:
$langle w,z rangle=sum_k=1^n langle w,v_kranglelangle v_k,zrangle$
inner-product-space
asked Sep 2 at 9:01
mathamity
639
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up vote
1
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Let $v_1,v_2,v_3,...,v_n$ be an orthonormal basis of $V$. Show that for any vectors $w$ and $z$ of $V$:
$langle w,z rangle=sum_k=1^n langle w,v_kranglelangle v_k,zrangle$
inner-product-space
asked Sep 2 at 9:01
mathamity
639
Hint: try to prove that $w = sum_i = 1^n langle w, v_irangle cdot v_i $.
– Xiangxiang Xu
Sep 2 at 9:09
@XiangxiangXu $w=sum_k=1^na_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=langle w,v_k rangle$
– mathamity
Sep 2 at 9:20
Then suppose $w = sum_i = 1^n a_i v_i, z = sum_i = 1^n b_i v_i$, can you show that $langle w, z rangle = sum_i = 1^n a_i b_i$?
– Xiangxiang Xu
Sep 2 at 9:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $v_1,v_2,v_3,...,v_n$ be an orthonormal basis of $V$. Show that for any vectors $w$ and $z$ of $V$:
$langle w,z rangle=sum_k=1^n langle w,v_kranglelangle v_k,zrangle$
inner-product-space
asked Sep 2 at 9:01
mathamity
639
Let $v_1,v_2,v_3,...,v_n$ be an orthonormal basis of $V$. Show that for any vectors $w$ and $z$ of $V$:
$langle w,z rangle=sum_k=1^n langle w,v_kranglelangle v_k,zrangle$
inner-product-space
inner-product-space
asked Sep 2 at 9:01
mathamity
639
asked Sep 2 at 9:01
mathamity
639
edited Sep 12 at 8:35
asked Sep 2 at 9:01
mathamity
639
asked Sep 2 at 9:01
mathamity
639
asked Sep 2 at 9:01
mathamity
639
639
Hint: try to prove that $w = sum_i = 1^n langle w, v_irangle cdot v_i $.
– Xiangxiang Xu
Sep 2 at 9:09
@XiangxiangXu $w=sum_k=1^na_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=langle w,v_k rangle$
– mathamity
Sep 2 at 9:20
Then suppose $w = sum_i = 1^n a_i v_i, z = sum_i = 1^n b_i v_i$, can you show that $langle w, z rangle = sum_i = 1^n a_i b_i$?
– Xiangxiang Xu
Sep 2 at 9:32
add a comment |Â
Hint: try to prove that $w = sum_i = 1^n langle w, v_irangle cdot v_i $.
– Xiangxiang Xu
Sep 2 at 9:09
@XiangxiangXu $w=sum_k=1^na_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=langle w,v_k rangle$
– mathamity
Sep 2 at 9:20
Then suppose $w = sum_i = 1^n a_i v_i, z = sum_i = 1^n b_i v_i$, can you show that $langle w, z rangle = sum_i = 1^n a_i b_i$?
– Xiangxiang Xu
Sep 2 at 9:32
Hint: try to prove that $w = sum_i = 1^n langle w, v_irangle cdot v_i $.
– Xiangxiang Xu
Sep 2 at 9:09
Hint: try to prove that $w = sum_i = 1^n langle w, v_irangle cdot v_i $.
– Xiangxiang Xu
Sep 2 at 9:09
@XiangxiangXu $w=sum_k=1^na_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=langle w,v_k rangle$
– mathamity
Sep 2 at 9:20
@XiangxiangXu $w=sum_k=1^na_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=langle w,v_k rangle$
– mathamity
Sep 2 at 9:20
Then suppose $w = sum_i = 1^n a_i v_i, z = sum_i = 1^n b_i v_i$, can you show that $langle w, z rangle = sum_i = 1^n a_i b_i$?
– Xiangxiang Xu
Sep 2 at 9:32
Then suppose $w = sum_i = 1^n a_i v_i, z = sum_i = 1^n b_i v_i$, can you show that $langle w, z rangle = sum_i = 1^n a_i b_i$?
– Xiangxiang Xu
Sep 2 at 9:32
add a comment |Â
2 Answers
2
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0
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We have that
- $w=sum_i=1^n a_i v_i$
- $z=sum_j=1^n b_j v_j$
then
$$langle w,z rangle=langlesum_i=1^n a_i v_i,sum_j=1^n b_j v_j
rangle=sum_k=1^n a_kb_k=sum_k=1^n a_kv_k^Tcdot v_kb_k=
sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 9:20
gimusi
72.2k73888
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
add a comment |Â
up vote
0
down vote
Since $v_1, ldots, v_n$ is an orthonormal basis, we have $langle v_k, v_jrangle = delta_kj$ and $x = sum_i=1^n langle x, v_irangle v_i$ for all $x in V$.
Therefore
$$langle w,z rangle= leftlangle sum_k=1^n langle w, v_krangle v_k, sum_j=1^n langle z, v_jrangle v_krightrangle = sum_k=1^nsum_j=1^n langle w,v_kranglelangle v_j,zrangle underbracelangle v_k, v_jrangle_delta_kj = sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 14:57
mechanodroid
24k52244
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have that
- $w=sum_i=1^n a_i v_i$
- $z=sum_j=1^n b_j v_j$
then
$$langle w,z rangle=langlesum_i=1^n a_i v_i,sum_j=1^n b_j v_j
rangle=sum_k=1^n a_kb_k=sum_k=1^n a_kv_k^Tcdot v_kb_k=
sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 9:20
gimusi
72.2k73888
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
add a comment |Â
up vote
0
down vote
We have that
- $w=sum_i=1^n a_i v_i$
- $z=sum_j=1^n b_j v_j$
then
$$langle w,z rangle=langlesum_i=1^n a_i v_i,sum_j=1^n b_j v_j
rangle=sum_k=1^n a_kb_k=sum_k=1^n a_kv_k^Tcdot v_kb_k=
sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 9:20
gimusi
72.2k73888
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have that
- $w=sum_i=1^n a_i v_i$
- $z=sum_j=1^n b_j v_j$
then
$$langle w,z rangle=langlesum_i=1^n a_i v_i,sum_j=1^n b_j v_j
rangle=sum_k=1^n a_kb_k=sum_k=1^n a_kv_k^Tcdot v_kb_k=
sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 9:20
gimusi
72.2k73888
We have that
- $w=sum_i=1^n a_i v_i$
- $z=sum_j=1^n b_j v_j$
then
$$langle w,z rangle=langlesum_i=1^n a_i v_i,sum_j=1^n b_j v_j
rangle=sum_k=1^n a_kb_k=sum_k=1^n a_kv_k^Tcdot v_kb_k=
sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 9:20
gimusi
72.2k73888
edited Sep 2 at 9:25
answered Sep 2 at 9:20
gimusi
72.2k73888
answered Sep 2 at 9:20
gimusi
72.2k73888
answered Sep 2 at 9:20
gimusi
72.2k73888
72.2k73888
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
add a comment |Â
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
you have used dot product definition. can we prove using only the generalised inner product definition?
– mathamity
Sep 2 at 9:23
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
@mathamity Yes the main steps are equivalent, that is $langle v_i,v_j rangle=0$ and $langle w,v_k rangle=a_k$.
– gimusi
Sep 2 at 9:25
add a comment |Â
up vote
0
down vote
Since $v_1, ldots, v_n$ is an orthonormal basis, we have $langle v_k, v_jrangle = delta_kj$ and $x = sum_i=1^n langle x, v_irangle v_i$ for all $x in V$.
Therefore
$$langle w,z rangle= leftlangle sum_k=1^n langle w, v_krangle v_k, sum_j=1^n langle z, v_jrangle v_krightrangle = sum_k=1^nsum_j=1^n langle w,v_kranglelangle v_j,zrangle underbracelangle v_k, v_jrangle_delta_kj = sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 14:57
mechanodroid
24k52244
add a comment |Â
up vote
0
down vote
Since $v_1, ldots, v_n$ is an orthonormal basis, we have $langle v_k, v_jrangle = delta_kj$ and $x = sum_i=1^n langle x, v_irangle v_i$ for all $x in V$.
Therefore
$$langle w,z rangle= leftlangle sum_k=1^n langle w, v_krangle v_k, sum_j=1^n langle z, v_jrangle v_krightrangle = sum_k=1^nsum_j=1^n langle w,v_kranglelangle v_j,zrangle underbracelangle v_k, v_jrangle_delta_kj = sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 14:57
mechanodroid
24k52244
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $v_1, ldots, v_n$ is an orthonormal basis, we have $langle v_k, v_jrangle = delta_kj$ and $x = sum_i=1^n langle x, v_irangle v_i$ for all $x in V$.
Therefore
$$langle w,z rangle= leftlangle sum_k=1^n langle w, v_krangle v_k, sum_j=1^n langle z, v_jrangle v_krightrangle = sum_k=1^nsum_j=1^n langle w,v_kranglelangle v_j,zrangle underbracelangle v_k, v_jrangle_delta_kj = sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 14:57
mechanodroid
24k52244
Since $v_1, ldots, v_n$ is an orthonormal basis, we have $langle v_k, v_jrangle = delta_kj$ and $x = sum_i=1^n langle x, v_irangle v_i$ for all $x in V$.
Therefore
$$langle w,z rangle= leftlangle sum_k=1^n langle w, v_krangle v_k, sum_j=1^n langle z, v_jrangle v_krightrangle = sum_k=1^nsum_j=1^n langle w,v_kranglelangle v_j,zrangle underbracelangle v_k, v_jrangle_delta_kj = sum_k=1^n langle w,v_kranglelangle v_k,zrangle$$
answered Sep 2 at 14:57
mechanodroid
24k52244
answered Sep 2 at 14:57
mechanodroid
24k52244
answered Sep 2 at 14:57
mechanodroid
24k52244
answered Sep 2 at 14:57
mechanodroid
24k52244
24k52244
add a comment |Â
add a comment |Â
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Hint: try to prove that $w = sum_i = 1^n langle w, v_irangle cdot v_i $.
– Xiangxiang Xu
Sep 2 at 9:09
@XiangxiangXu $w=sum_k=1^na_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=langle w,v_k rangle$
– mathamity
Sep 2 at 9:20
Then suppose $w = sum_i = 1^n a_i v_i, z = sum_i = 1^n b_i v_i$, can you show that $langle w, z rangle = sum_i = 1^n a_i b_i$?
– Xiangxiang Xu
Sep 2 at 9:32