Mixing
Decomposing Square Matrix Into Two Matrices Which Are Transposes of Each Other
Clash Royale CLAN TAG#URR8PPP
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Let $X$ be a real-valued square $n times n$ matrix.
Is there decomposition $X = Lambda Lambda^top$ where $Lambda$ is a a real-valued $n times k$ matrix, that always exists?
linear-algebra matrices matrix-decomposition
asked Jul 23 at 18:24
stollenm
21519
add a comment |Â
up vote
0
down vote
favorite
Let $X$ be a real-valued square $n times n$ matrix.
Is there decomposition $X = Lambda Lambda^top$ where $Lambda$ is a a real-valued $n times k$ matrix, that always exists?
linear-algebra matrices matrix-decomposition
asked Jul 23 at 18:24
stollenm
21519
3
$Lambda Lambda^top$ is always symmetric.
– Lord Shark the Unknown
Jul 23 at 18:27
3
..and positive semi-definite
– mheldman
Jul 23 at 18:32
Oh of course, Thanks!
– stollenm
Jul 23 at 18:32
Interestingly, if $X$ is complex and symmetric, then there exists a complex $Lambda$ so that $X = Lambda Lambda^T$
– Omnomnomnom
Jul 23 at 18:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a real-valued square $n times n$ matrix.
Is there decomposition $X = Lambda Lambda^top$ where $Lambda$ is a a real-valued $n times k$ matrix, that always exists?
linear-algebra matrices matrix-decomposition
asked Jul 23 at 18:24
stollenm
21519
Let $X$ be a real-valued square $n times n$ matrix.
Is there decomposition $X = Lambda Lambda^top$ where $Lambda$ is a a real-valued $n times k$ matrix, that always exists?
linear-algebra matrices matrix-decomposition
asked Jul 23 at 18:24
stollenm
21519
asked Jul 23 at 18:24
stollenm
21519
asked Jul 23 at 18:24
stollenm
21519
asked Jul 23 at 18:24
stollenm
21519
21519
3
$Lambda Lambda^top$ is always symmetric.
– Lord Shark the Unknown
Jul 23 at 18:27
3
..and positive semi-definite
– mheldman
Jul 23 at 18:32
Oh of course, Thanks!
– stollenm
Jul 23 at 18:32
Interestingly, if $X$ is complex and symmetric, then there exists a complex $Lambda$ so that $X = Lambda Lambda^T$
– Omnomnomnom
Jul 23 at 18:43
add a comment |Â
3
$Lambda Lambda^top$ is always symmetric.
– Lord Shark the Unknown
Jul 23 at 18:27
3
..and positive semi-definite
– mheldman
Jul 23 at 18:32
Oh of course, Thanks!
– stollenm
Jul 23 at 18:32
Interestingly, if $X$ is complex and symmetric, then there exists a complex $Lambda$ so that $X = Lambda Lambda^T$
– Omnomnomnom
Jul 23 at 18:43
3
3
$Lambda Lambda^top$ is always symmetric.
– Lord Shark the Unknown
Jul 23 at 18:27
$Lambda Lambda^top$ is always symmetric.
– Lord Shark the Unknown
Jul 23 at 18:27
3
3
..and positive semi-definite
– mheldman
Jul 23 at 18:32
..and positive semi-definite
– mheldman
Jul 23 at 18:32
Oh of course, Thanks!
– stollenm
Jul 23 at 18:32
Oh of course, Thanks!
– stollenm
Jul 23 at 18:32
Interestingly, if $X$ is complex and symmetric, then there exists a complex $Lambda$ so that $X = Lambda Lambda^T$
– Omnomnomnom
Jul 23 at 18:43
Interestingly, if $X$ is complex and symmetric, then there exists a complex $Lambda$ so that $X = Lambda Lambda^T$
– Omnomnomnom
Jul 23 at 18:43
add a comment |Â
2 Answers
2
active
oldest
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1
down vote
accepted
Every real symmetric matrix can be written in the form
$$ X = Q D Q^T $$
where $Q$ is formed from an orthonormal set of eigenvectors of $X$ and the diagonal of $D$ contains the corresponding eigenvalues. If the eigenvalues are non-negative, then the real matrix $Lambda = PD^1/2$ satisfies your condition.
$$ X = (PD^1/2)(D^1/2P^T) = Lambda Lambda^T $$
Note that all matrices formed by the product $ Lambda Lambda^T $ are positive semidefinite, so the following statements are equivalent:
- $X$ is a real positive semidefinite matrix.
- There exists a real matrix $Lambda$ such that $X = Lambda Lambda^T$.
answered Jul 23 at 19:10
Jared Goguen
890514
add a comment |Â
up vote
1
down vote
If $X in mathbbC^m times m$ is positive definite and hermitian then there exists the Cholesky decomposition such that
$$ X = R^*R$$
where
$$ X = underbraceR_1^*R_2^* cdots R_m^* _R^*underbraceR_m cdots R_2 R_1_R$$
with
$$ X = R^*R , r_jj > 0 $$
which can be done as
answered Jul 23 at 19:22
RHowe
1,010815
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Every real symmetric matrix can be written in the form
$$ X = Q D Q^T $$
where $Q$ is formed from an orthonormal set of eigenvectors of $X$ and the diagonal of $D$ contains the corresponding eigenvalues. If the eigenvalues are non-negative, then the real matrix $Lambda = PD^1/2$ satisfies your condition.
$$ X = (PD^1/2)(D^1/2P^T) = Lambda Lambda^T $$
Note that all matrices formed by the product $ Lambda Lambda^T $ are positive semidefinite, so the following statements are equivalent:
- $X$ is a real positive semidefinite matrix.
- There exists a real matrix $Lambda$ such that $X = Lambda Lambda^T$.
answered Jul 23 at 19:10
Jared Goguen
890514
add a comment |Â
up vote
1
down vote
accepted
Every real symmetric matrix can be written in the form
$$ X = Q D Q^T $$
where $Q$ is formed from an orthonormal set of eigenvectors of $X$ and the diagonal of $D$ contains the corresponding eigenvalues. If the eigenvalues are non-negative, then the real matrix $Lambda = PD^1/2$ satisfies your condition.
$$ X = (PD^1/2)(D^1/2P^T) = Lambda Lambda^T $$
Note that all matrices formed by the product $ Lambda Lambda^T $ are positive semidefinite, so the following statements are equivalent:
- $X$ is a real positive semidefinite matrix.
- There exists a real matrix $Lambda$ such that $X = Lambda Lambda^T$.
answered Jul 23 at 19:10
Jared Goguen
890514
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Every real symmetric matrix can be written in the form
$$ X = Q D Q^T $$
where $Q$ is formed from an orthonormal set of eigenvectors of $X$ and the diagonal of $D$ contains the corresponding eigenvalues. If the eigenvalues are non-negative, then the real matrix $Lambda = PD^1/2$ satisfies your condition.
$$ X = (PD^1/2)(D^1/2P^T) = Lambda Lambda^T $$
Note that all matrices formed by the product $ Lambda Lambda^T $ are positive semidefinite, so the following statements are equivalent:
- $X$ is a real positive semidefinite matrix.
- There exists a real matrix $Lambda$ such that $X = Lambda Lambda^T$.
answered Jul 23 at 19:10
Jared Goguen
890514
Every real symmetric matrix can be written in the form
$$ X = Q D Q^T $$
where $Q$ is formed from an orthonormal set of eigenvectors of $X$ and the diagonal of $D$ contains the corresponding eigenvalues. If the eigenvalues are non-negative, then the real matrix $Lambda = PD^1/2$ satisfies your condition.
$$ X = (PD^1/2)(D^1/2P^T) = Lambda Lambda^T $$
Note that all matrices formed by the product $ Lambda Lambda^T $ are positive semidefinite, so the following statements are equivalent:
- $X$ is a real positive semidefinite matrix.
- There exists a real matrix $Lambda$ such that $X = Lambda Lambda^T$.
answered Jul 23 at 19:10
Jared Goguen
890514
answered Jul 23 at 19:10
Jared Goguen
890514
answered Jul 23 at 19:10
Jared Goguen
890514
answered Jul 23 at 19:10
Jared Goguen
890514
890514
add a comment |Â
add a comment |Â
up vote
1
down vote
If $X in mathbbC^m times m$ is positive definite and hermitian then there exists the Cholesky decomposition such that
$$ X = R^*R$$
where
$$ X = underbraceR_1^*R_2^* cdots R_m^* _R^*underbraceR_m cdots R_2 R_1_R$$
with
$$ X = R^*R , r_jj > 0 $$
which can be done as
answered Jul 23 at 19:22
RHowe
1,010815
add a comment |Â
up vote
1
down vote
If $X in mathbbC^m times m$ is positive definite and hermitian then there exists the Cholesky decomposition such that
$$ X = R^*R$$
where
$$ X = underbraceR_1^*R_2^* cdots R_m^* _R^*underbraceR_m cdots R_2 R_1_R$$
with
$$ X = R^*R , r_jj > 0 $$
which can be done as
answered Jul 23 at 19:22
RHowe
1,010815
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $X in mathbbC^m times m$ is positive definite and hermitian then there exists the Cholesky decomposition such that
$$ X = R^*R$$
where
$$ X = underbraceR_1^*R_2^* cdots R_m^* _R^*underbraceR_m cdots R_2 R_1_R$$
with
$$ X = R^*R , r_jj > 0 $$
which can be done as
answered Jul 23 at 19:22
RHowe
1,010815
If $X in mathbbC^m times m$ is positive definite and hermitian then there exists the Cholesky decomposition such that
$$ X = R^*R$$
where
$$ X = underbraceR_1^*R_2^* cdots R_m^* _R^*underbraceR_m cdots R_2 R_1_R$$
with
$$ X = R^*R , r_jj > 0 $$
which can be done as
answered Jul 23 at 19:22
RHowe
1,010815
edited Jul 23 at 22:55
answered Jul 23 at 19:22
RHowe
1,010815
answered Jul 23 at 19:22
RHowe
1,010815
answered Jul 23 at 19:22
RHowe
1,010815
1,010815
add a comment |Â
add a comment |Â
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3
$Lambda Lambda^top$ is always symmetric.
– Lord Shark the Unknown
Jul 23 at 18:27
3
..and positive semi-definite
– mheldman
Jul 23 at 18:32
Oh of course, Thanks!
– stollenm
Jul 23 at 18:32
Interestingly, if $X$ is complex and symmetric, then there exists a complex $Lambda$ so that $X = Lambda Lambda^T$
– Omnomnomnom
Jul 23 at 18:43