Sum of product of orthonormal vectors [closed]
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If $v_1,v_2,...v_n$ are orthonormal vectors in $in mathbb R^n$ is there anything special about $v_1v_1^T+v_2v_2^T+...+v_nv_n^T$
linear-algebra
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
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closed as off-topic by Saad, Brahadeesh, Shaun, stressed out, user10354138 Dec 20 '18 at 17:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Shaun, stressed out, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $v_1,v_2,...v_n$ are orthonormal vectors in $in mathbb R^n$ is there anything special about $v_1v_1^T+v_2v_2^T+...+v_nv_n^T$
linear-algebra
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
$endgroup$
closed as off-topic by Saad, Brahadeesh, Shaun, stressed out, user10354138 Dec 20 '18 at 17:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Shaun, stressed out, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Row vectors or column vectors?
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– Kavi Rama Murthy
Dec 20 '18 at 6:06
1
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@KaviRamaMurthy column vectors. I guess I have found the answer. It looks like Identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:06
$begingroup$
Orthogonal or orthonormal? Let $V$ be the matrix, then $V v_k = |v_k|^2 v_k$.
$endgroup$
– copper.hat
Dec 20 '18 at 6:07
1
$begingroup$
@copper.hat they have unit length and orthogonal to each other, so yes they are orthonormal
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:08
3
$begingroup$
Then it is the identity since it agrees with the identity on a basis.
$endgroup$
– copper.hat
Dec 20 '18 at 6:08
add a comment |
$begingroup$
If $v_1,v_2,...v_n$ are orthonormal vectors in $in mathbb R^n$ is there anything special about $v_1v_1^T+v_2v_2^T+...+v_nv_n^T$
linear-algebra
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
$endgroup$
If $v_1,v_2,...v_n$ are orthonormal vectors in $in mathbb R^n$ is there anything special about $v_1v_1^T+v_2v_2^T+...+v_nv_n^T$
linear-algebra
linear-algebra
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
edited Dec 20 '18 at 6:08
Studying Optimization
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
asked Dec 20 '18 at 6:04
Studying OptimizationStudying Optimization
917
917
closed as off-topic by Saad, Brahadeesh, Shaun, stressed out, user10354138 Dec 20 '18 at 17:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Shaun, stressed out, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Brahadeesh, Shaun, stressed out, user10354138 Dec 20 '18 at 17:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Shaun, stressed out, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Row vectors or column vectors?
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:06
1
$begingroup$
@KaviRamaMurthy column vectors. I guess I have found the answer. It looks like Identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:06
$begingroup$
Orthogonal or orthonormal? Let $V$ be the matrix, then $V v_k = |v_k|^2 v_k$.
$endgroup$
– copper.hat
Dec 20 '18 at 6:07
1
$begingroup$
@copper.hat they have unit length and orthogonal to each other, so yes they are orthonormal
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:08
3
$begingroup$
Then it is the identity since it agrees with the identity on a basis.
$endgroup$
– copper.hat
Dec 20 '18 at 6:08
add a comment |
$begingroup$
Row vectors or column vectors?
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:06
1
$begingroup$
@KaviRamaMurthy column vectors. I guess I have found the answer. It looks like Identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:06
$begingroup$
Orthogonal or orthonormal? Let $V$ be the matrix, then $V v_k = |v_k|^2 v_k$.
$endgroup$
– copper.hat
Dec 20 '18 at 6:07
1
$begingroup$
@copper.hat they have unit length and orthogonal to each other, so yes they are orthonormal
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:08
3
$begingroup$
Then it is the identity since it agrees with the identity on a basis.
$endgroup$
– copper.hat
Dec 20 '18 at 6:08
$begingroup$
Row vectors or column vectors?
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:06
$begingroup$
Row vectors or column vectors?
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:06
1
1
$begingroup$
@KaviRamaMurthy column vectors. I guess I have found the answer. It looks like Identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:06
$begingroup$
@KaviRamaMurthy column vectors. I guess I have found the answer. It looks like Identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:06
$begingroup$
Orthogonal or orthonormal? Let $V$ be the matrix, then $V v_k = |v_k|^2 v_k$.
$endgroup$
– copper.hat
Dec 20 '18 at 6:07
$begingroup$
Orthogonal or orthonormal? Let $V$ be the matrix, then $V v_k = |v_k|^2 v_k$.
$endgroup$
– copper.hat
Dec 20 '18 at 6:07
1
1
$begingroup$
@copper.hat they have unit length and orthogonal to each other, so yes they are orthonormal
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:08
$begingroup$
@copper.hat they have unit length and orthogonal to each other, so yes they are orthonormal
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:08
3
3
$begingroup$
Then it is the identity since it agrees with the identity on a basis.
$endgroup$
– copper.hat
Dec 20 '18 at 6:08
$begingroup$
Then it is the identity since it agrees with the identity on a basis.
$endgroup$
– copper.hat
Dec 20 '18 at 6:08
add a comment |
3 Answers
3
active
oldest
votes
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If $V = [v_1 ldots v_n]$, we have that $V^T V = I$. This means $V^T = V^{-1}$, but $VV^{-1} = I$ so, $VV^T = I$. Expanding the latter, we get your desired result.
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
We have $v_j v_j^T=||v_j||^2=1$ for all $j=1,2,...,n.$ Hence $v_1v_1^T+v_2v_2^T+...+v_nv_n^T=n$.
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
$endgroup$
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
add a comment |
$begingroup$
If $v_i$ is a column vector then multiply $v_1v_1^T+dots+v_nv_n^T$ by any vector $x$:
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)
$$
Since $v_i$ make an orthonormal basis, the $v_i^Tx$ is the coefficient in the decomposition of $x$:
$x=a_1v_1+dots+a_nv_n$ implies $v_i^Tx=v_i^T(a_1v_1+dots+a_nv_n)=a_iv_i^Tv_i=a_i$. Therefore
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)=x
$$
which implies that $v_1v_1^T+dots+v_nv_n^T$ is the identity matrix.
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $V = [v_1 ldots v_n]$, we have that $V^T V = I$. This means $V^T = V^{-1}$, but $VV^{-1} = I$ so, $VV^T = I$. Expanding the latter, we get your desired result.
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
If $V = [v_1 ldots v_n]$, we have that $V^T V = I$. This means $V^T = V^{-1}$, but $VV^{-1} = I$ so, $VV^T = I$. Expanding the latter, we get your desired result.
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
If $V = [v_1 ldots v_n]$, we have that $V^T V = I$. This means $V^T = V^{-1}$, but $VV^{-1} = I$ so, $VV^T = I$. Expanding the latter, we get your desired result.
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
If $V = [v_1 ldots v_n]$, we have that $V^T V = I$. This means $V^T = V^{-1}$, but $VV^{-1} = I$ so, $VV^T = I$. Expanding the latter, we get your desired result.
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
answered Dec 20 '18 at 7:21
Ahmad BazziAhmad Bazzi
8,3622824
8,3622824
add a comment |
add a comment |
$begingroup$
We have $v_j v_j^T=||v_j||^2=1$ for all $j=1,2,...,n.$ Hence $v_1v_1^T+v_2v_2^T+...+v_nv_n^T=n$.
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
$endgroup$
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
add a comment |
$begingroup$
We have $v_j v_j^T=||v_j||^2=1$ for all $j=1,2,...,n.$ Hence $v_1v_1^T+v_2v_2^T+...+v_nv_n^T=n$.
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
$endgroup$
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
add a comment |
$begingroup$
We have $v_j v_j^T=||v_j||^2=1$ for all $j=1,2,...,n.$ Hence $v_1v_1^T+v_2v_2^T+...+v_nv_n^T=n$.
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
$endgroup$
We have $v_j v_j^T=||v_j||^2=1$ for all $j=1,2,...,n.$ Hence $v_1v_1^T+v_2v_2^T+...+v_nv_n^T=n$.
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
answered Dec 20 '18 at 6:19
FredFred
47.2k1849
47.2k1849
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
add a comment |
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
I believe the vectors are column vectors so the terms are matrices, not scalars.
$endgroup$
– John Douma
Dec 20 '18 at 6:25
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
$begingroup$
Yes, it is an identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:56
add a comment |
$begingroup$
If $v_i$ is a column vector then multiply $v_1v_1^T+dots+v_nv_n^T$ by any vector $x$:
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)
$$
Since $v_i$ make an orthonormal basis, the $v_i^Tx$ is the coefficient in the decomposition of $x$:
$x=a_1v_1+dots+a_nv_n$ implies $v_i^Tx=v_i^T(a_1v_1+dots+a_nv_n)=a_iv_i^Tv_i=a_i$. Therefore
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)=x
$$
which implies that $v_1v_1^T+dots+v_nv_n^T$ is the identity matrix.
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
$endgroup$
add a comment |
$begingroup$
If $v_i$ is a column vector then multiply $v_1v_1^T+dots+v_nv_n^T$ by any vector $x$:
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)
$$
Since $v_i$ make an orthonormal basis, the $v_i^Tx$ is the coefficient in the decomposition of $x$:
$x=a_1v_1+dots+a_nv_n$ implies $v_i^Tx=v_i^T(a_1v_1+dots+a_nv_n)=a_iv_i^Tv_i=a_i$. Therefore
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)=x
$$
which implies that $v_1v_1^T+dots+v_nv_n^T$ is the identity matrix.
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
$endgroup$
add a comment |
$begingroup$
If $v_i$ is a column vector then multiply $v_1v_1^T+dots+v_nv_n^T$ by any vector $x$:
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)
$$
Since $v_i$ make an orthonormal basis, the $v_i^Tx$ is the coefficient in the decomposition of $x$:
$x=a_1v_1+dots+a_nv_n$ implies $v_i^Tx=v_i^T(a_1v_1+dots+a_nv_n)=a_iv_i^Tv_i=a_i$. Therefore
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)=x
$$
which implies that $v_1v_1^T+dots+v_nv_n^T$ is the identity matrix.
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
$endgroup$
If $v_i$ is a column vector then multiply $v_1v_1^T+dots+v_nv_n^T$ by any vector $x$:
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)
$$
Since $v_i$ make an orthonormal basis, the $v_i^Tx$ is the coefficient in the decomposition of $x$:
$x=a_1v_1+dots+a_nv_n$ implies $v_i^Tx=v_i^T(a_1v_1+dots+a_nv_n)=a_iv_i^Tv_i=a_i$. Therefore
$$
(v_1v_1^T+dots+v_nv_n^T)x=v_1(v_1^Tx)+dots+v_n(v_n^Tx)=x
$$
which implies that $v_1v_1^T+dots+v_nv_n^T$ is the identity matrix.
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
answered Dec 20 '18 at 7:13
Sergei GolovanSergei Golovan
1,393147
1,393147
add a comment |
add a comment |
$begingroup$
Row vectors or column vectors?
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:06
1
$begingroup$
@KaviRamaMurthy column vectors. I guess I have found the answer. It looks like Identity matrix
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:06
$begingroup$
Orthogonal or orthonormal? Let $V$ be the matrix, then $V v_k = |v_k|^2 v_k$.
$endgroup$
– copper.hat
Dec 20 '18 at 6:07
1
$begingroup$
@copper.hat they have unit length and orthogonal to each other, so yes they are orthonormal
$endgroup$
– Studying Optimization
Dec 20 '18 at 6:08
3
$begingroup$
Then it is the identity since it agrees with the identity on a basis.
$endgroup$
– copper.hat
Dec 20 '18 at 6:08