Mixing
The existence of an orthonormal matrix
Let $U$ be an orthonormal matrix. The dimension of $U$ is equal to $n times n$. The first row of $U$ has the following form:
$$bigg( frac{1}{sqrt{n}}, ldots , frac{1}{sqrt{n}} bigg). tag{1}$$
How can I prove that $U$ with property (1) exists?
linear-algebra
asked Nov 20 '18 at 17:40
Hendrra
1,079516
add a comment |
Let $U$ be an orthonormal matrix. The dimension of $U$ is equal to $n times n$. The first row of $U$ has the following form:
$$bigg( frac{1}{sqrt{n}}, ldots , frac{1}{sqrt{n}} bigg). tag{1}$$
How can I prove that $U$ with property (1) exists?
linear-algebra
asked Nov 20 '18 at 17:40
Hendrra
1,079516
add a comment |
Let $U$ be an orthonormal matrix. The dimension of $U$ is equal to $n times n$. The first row of $U$ has the following form:
$$bigg( frac{1}{sqrt{n}}, ldots , frac{1}{sqrt{n}} bigg). tag{1}$$
How can I prove that $U$ with property (1) exists?
linear-algebra
asked Nov 20 '18 at 17:40
Hendrra
1,079516
Let $U$ be an orthonormal matrix. The dimension of $U$ is equal to $n times n$. The first row of $U$ has the following form:
$$bigg( frac{1}{sqrt{n}}, ldots , frac{1}{sqrt{n}} bigg). tag{1}$$
How can I prove that $U$ with property (1) exists?
linear-algebra
linear-algebra
asked Nov 20 '18 at 17:40
Hendrra
1,079516
asked Nov 20 '18 at 17:40
Hendrra
1,079516
asked Nov 20 '18 at 17:40
Hendrra
1,079516
asked Nov 20 '18 at 17:40
Hendrra
1,079516
asked Nov 20 '18 at 17:40
Hendrra
1,079516
1,079516
add a comment |
add a comment |
3 Answers
3
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oldest
votes
Take your row vector (call it $v_1$) and extend this set to a basis of $mathbb{R}^n$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$v_1, dots, v_n$}. Then, the matrix U such that its jth row is $v_j$ is an orthonormal matrix.
answered Nov 20 '18 at 17:56
M. Santos
764
add a comment |
$u:=(frac{1}{sqrt{n}},ldots,frac{1}{sqrt{n}})$ is a unit vector in your vector space. You can extend it to a basis ${u,v_{2},ldots,v_{n}}$ of $V$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: ${e_{1},e_{2},ldots,e_{n}}$ of $V$, where $e_{1}=u$. The matrix with rows $e_{1},ldots e_{n}$ will then be a unitary matrix.
answered Nov 20 '18 at 17:52
ervx
10.3k31338
add a comment |
Consider the vectors $v_1=e_1+dots+e_n$, $v_i=e_i$ for $i=2,dots,n$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $(e_1+dots+e_n)/sqrt n$, which is precisely the vector you want.
answered Nov 20 '18 at 17:51
Federico
4,709514
1
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Take your row vector (call it $v_1$) and extend this set to a basis of $mathbb{R}^n$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$v_1, dots, v_n$}. Then, the matrix U such that its jth row is $v_j$ is an orthonormal matrix.
answered Nov 20 '18 at 17:56
M. Santos
764
add a comment |
Take your row vector (call it $v_1$) and extend this set to a basis of $mathbb{R}^n$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$v_1, dots, v_n$}. Then, the matrix U such that its jth row is $v_j$ is an orthonormal matrix.
answered Nov 20 '18 at 17:56
M. Santos
764
add a comment |
Take your row vector (call it $v_1$) and extend this set to a basis of $mathbb{R}^n$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$v_1, dots, v_n$}. Then, the matrix U such that its jth row is $v_j$ is an orthonormal matrix.
answered Nov 20 '18 at 17:56
M. Santos
764
Take your row vector (call it $v_1$) and extend this set to a basis of $mathbb{R}^n$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$v_1, dots, v_n$}. Then, the matrix U such that its jth row is $v_j$ is an orthonormal matrix.
answered Nov 20 '18 at 17:56
M. Santos
764
answered Nov 20 '18 at 17:56
M. Santos
764
answered Nov 20 '18 at 17:56
M. Santos
764
answered Nov 20 '18 at 17:56
M. Santos
764
764
add a comment |
add a comment |
$u:=(frac{1}{sqrt{n}},ldots,frac{1}{sqrt{n}})$ is a unit vector in your vector space. You can extend it to a basis ${u,v_{2},ldots,v_{n}}$ of $V$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: ${e_{1},e_{2},ldots,e_{n}}$ of $V$, where $e_{1}=u$. The matrix with rows $e_{1},ldots e_{n}$ will then be a unitary matrix.
answered Nov 20 '18 at 17:52
ervx
10.3k31338
add a comment |
$u:=(frac{1}{sqrt{n}},ldots,frac{1}{sqrt{n}})$ is a unit vector in your vector space. You can extend it to a basis ${u,v_{2},ldots,v_{n}}$ of $V$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: ${e_{1},e_{2},ldots,e_{n}}$ of $V$, where $e_{1}=u$. The matrix with rows $e_{1},ldots e_{n}$ will then be a unitary matrix.
answered Nov 20 '18 at 17:52
ervx
10.3k31338
add a comment |
$u:=(frac{1}{sqrt{n}},ldots,frac{1}{sqrt{n}})$ is a unit vector in your vector space. You can extend it to a basis ${u,v_{2},ldots,v_{n}}$ of $V$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: ${e_{1},e_{2},ldots,e_{n}}$ of $V$, where $e_{1}=u$. The matrix with rows $e_{1},ldots e_{n}$ will then be a unitary matrix.
answered Nov 20 '18 at 17:52
ervx
10.3k31338
$u:=(frac{1}{sqrt{n}},ldots,frac{1}{sqrt{n}})$ is a unit vector in your vector space. You can extend it to a basis ${u,v_{2},ldots,v_{n}}$ of $V$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: ${e_{1},e_{2},ldots,e_{n}}$ of $V$, where $e_{1}=u$. The matrix with rows $e_{1},ldots e_{n}$ will then be a unitary matrix.
answered Nov 20 '18 at 17:52
ervx
10.3k31338
answered Nov 20 '18 at 17:52
ervx
10.3k31338
answered Nov 20 '18 at 17:52
ervx
10.3k31338
answered Nov 20 '18 at 17:52
ervx
10.3k31338
10.3k31338
add a comment |
add a comment |
Consider the vectors $v_1=e_1+dots+e_n$, $v_i=e_i$ for $i=2,dots,n$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $(e_1+dots+e_n)/sqrt n$, which is precisely the vector you want.
answered Nov 20 '18 at 17:51
Federico
4,709514
1
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
add a comment |
Consider the vectors $v_1=e_1+dots+e_n$, $v_i=e_i$ for $i=2,dots,n$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $(e_1+dots+e_n)/sqrt n$, which is precisely the vector you want.
answered Nov 20 '18 at 17:51
Federico
4,709514
1
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
add a comment |
Consider the vectors $v_1=e_1+dots+e_n$, $v_i=e_i$ for $i=2,dots,n$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $(e_1+dots+e_n)/sqrt n$, which is precisely the vector you want.
answered Nov 20 '18 at 17:51
Federico
4,709514
Consider the vectors $v_1=e_1+dots+e_n$, $v_i=e_i$ for $i=2,dots,n$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $(e_1+dots+e_n)/sqrt n$, which is precisely the vector you want.
answered Nov 20 '18 at 17:51
Federico
4,709514
answered Nov 20 '18 at 17:51
Federico
4,709514
answered Nov 20 '18 at 17:51
Federico
4,709514
answered Nov 20 '18 at 17:51
Federico
4,709514
4,709514
1
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
add a comment |
1
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
1
1
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
Can the downvoter explain please? This answer is correct and was also the first to be posted
– Federico
Nov 20 '18 at 18:15
add a comment |
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