$xx^T$ is an extreme vector of positive semi-definite ($PSD_n$) matrices.
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Let $x in Bbb R^n$ be a vector.
Claim: $xx^T$ is an extreme vector of positive semi-definite ($PSD_n$) matrices.
We have to show that if $xx^T = A + B$, where $A,B in PSD_n$, then both $A,B$ are scalar multiple of $xx^T$.
$ns(A) cap ns(B) = nsxx^T = x^perp$, where ns means null-space.
So, either $ns(A) = x^perp textor Bbb R^n$. In case of $ns(A) = Bbb R^n$, we are done. I am having problem when $ns(A) = x^perp$, how to conclude that $A$ is scalar multiple of $xx^T$.
linear-algebra matrices positive-semidefinite
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 10:02
user8795
5,25761842
add a comment |Â
up vote
1
down vote
favorite
Let $x in Bbb R^n$ be a vector.
Claim: $xx^T$ is an extreme vector of positive semi-definite ($PSD_n$) matrices.
We have to show that if $xx^T = A + B$, where $A,B in PSD_n$, then both $A,B$ are scalar multiple of $xx^T$.
$ns(A) cap ns(B) = nsxx^T = x^perp$, where ns means null-space.
So, either $ns(A) = x^perp textor Bbb R^n$. In case of $ns(A) = Bbb R^n$, we are done. I am having problem when $ns(A) = x^perp$, how to conclude that $A$ is scalar multiple of $xx^T$.
linear-algebra matrices positive-semidefinite
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 10:02
user8795
5,25761842
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $x in Bbb R^n$ be a vector.
Claim: $xx^T$ is an extreme vector of positive semi-definite ($PSD_n$) matrices.
We have to show that if $xx^T = A + B$, where $A,B in PSD_n$, then both $A,B$ are scalar multiple of $xx^T$.
$ns(A) cap ns(B) = nsxx^T = x^perp$, where ns means null-space.
So, either $ns(A) = x^perp textor Bbb R^n$. In case of $ns(A) = Bbb R^n$, we are done. I am having problem when $ns(A) = x^perp$, how to conclude that $A$ is scalar multiple of $xx^T$.
linear-algebra matrices positive-semidefinite
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 10:02
user8795
5,25761842
Let $x in Bbb R^n$ be a vector.
Claim: $xx^T$ is an extreme vector of positive semi-definite ($PSD_n$) matrices.
We have to show that if $xx^T = A + B$, where $A,B in PSD_n$, then both $A,B$ are scalar multiple of $xx^T$.
$ns(A) cap ns(B) = nsxx^T = x^perp$, where ns means null-space.
So, either $ns(A) = x^perp textor Bbb R^n$. In case of $ns(A) = Bbb R^n$, we are done. I am having problem when $ns(A) = x^perp$, how to conclude that $A$ is scalar multiple of $xx^T$.
linear-algebra matrices positive-semidefinite
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
asked Aug 13 at 10:02
user8795
5,25761842
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
edited Aug 13 at 10:13
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Aug 13 at 10:02
user8795
5,25761842
asked Aug 13 at 10:02
user8795
5,25761842
asked Aug 13 at 10:02
user8795
5,25761842
5,25761842
add a comment |Â
add a comment |Â
1 Answer
1
active
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votes
up vote
1
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accepted
Suppose that $ns(A) = x^perp$. By the rank-nullity theorem, $A$ is a rank $1$ matrix. That is, we necessarily have $A = uv^T$ for some vectors $u,v$. Because $A$ is symmetric, $u$ is a multiple of $v$. Because $ns(A) = x^perp$, $v$ must be a multiple of $x$.
Conclude that $A$ is a multiple of $xx^T$, as desired.
answered Aug 13 at 12:42
Omnomnomnom
122k784170
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose that $ns(A) = x^perp$. By the rank-nullity theorem, $A$ is a rank $1$ matrix. That is, we necessarily have $A = uv^T$ for some vectors $u,v$. Because $A$ is symmetric, $u$ is a multiple of $v$. Because $ns(A) = x^perp$, $v$ must be a multiple of $x$.
Conclude that $A$ is a multiple of $xx^T$, as desired.
answered Aug 13 at 12:42
Omnomnomnom
122k784170
add a comment |Â
up vote
1
down vote
accepted
Suppose that $ns(A) = x^perp$. By the rank-nullity theorem, $A$ is a rank $1$ matrix. That is, we necessarily have $A = uv^T$ for some vectors $u,v$. Because $A$ is symmetric, $u$ is a multiple of $v$. Because $ns(A) = x^perp$, $v$ must be a multiple of $x$.
Conclude that $A$ is a multiple of $xx^T$, as desired.
answered Aug 13 at 12:42
Omnomnomnom
122k784170
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose that $ns(A) = x^perp$. By the rank-nullity theorem, $A$ is a rank $1$ matrix. That is, we necessarily have $A = uv^T$ for some vectors $u,v$. Because $A$ is symmetric, $u$ is a multiple of $v$. Because $ns(A) = x^perp$, $v$ must be a multiple of $x$.
Conclude that $A$ is a multiple of $xx^T$, as desired.
answered Aug 13 at 12:42
Omnomnomnom
122k784170
Suppose that $ns(A) = x^perp$. By the rank-nullity theorem, $A$ is a rank $1$ matrix. That is, we necessarily have $A = uv^T$ for some vectors $u,v$. Because $A$ is symmetric, $u$ is a multiple of $v$. Because $ns(A) = x^perp$, $v$ must be a multiple of $x$.
Conclude that $A$ is a multiple of $xx^T$, as desired.
answered Aug 13 at 12:42
Omnomnomnom
122k784170
answered Aug 13 at 12:42
Omnomnomnom
122k784170
answered Aug 13 at 12:42
Omnomnomnom
122k784170
answered Aug 13 at 12:42
Omnomnomnom
122k784170
122k784170
add a comment |Â
add a comment |Â
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