Positive Semidefinite Function
What does it mean for a function to be positive (semi)definite?
Assume we have a positive (semi)definite function $K(X, Y)$ that lives in $R^n$, and maps vector pairs in $R^n$ into a scalar in R. For instance, a kernel function is defined as $K(x,y) = <phi(x),phi(y)>$ for some appropriate mapping function $phi$. Assume further that we have n vectors, $v_1, cdots, v_n$, with real entries. We say that a matrix, A, has the entries $A_{ij} = K(v_i,v_j) = A_{ji} = K(v_j,v_i)$. Why we can conclude that A is a positive (semi)definite matrix?
Why the sum of positive (semi)definite functions is also positive (semi)definite?
linear-algebra functional-analysis covariance positive-semidefinite
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
asked Oct 20 '15 at 20:40
Jack Shi
617
add a comment |
What does it mean for a function to be positive (semi)definite?
Assume we have a positive (semi)definite function $K(X, Y)$ that lives in $R^n$, and maps vector pairs in $R^n$ into a scalar in R. For instance, a kernel function is defined as $K(x,y) = <phi(x),phi(y)>$ for some appropriate mapping function $phi$. Assume further that we have n vectors, $v_1, cdots, v_n$, with real entries. We say that a matrix, A, has the entries $A_{ij} = K(v_i,v_j) = A_{ji} = K(v_j,v_i)$. Why we can conclude that A is a positive (semi)definite matrix?
Why the sum of positive (semi)definite functions is also positive (semi)definite?
linear-algebra functional-analysis covariance positive-semidefinite
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
asked Oct 20 '15 at 20:40
Jack Shi
617
The question asks when a "function" is positive definite and then talks about bilinear forms, so I assume that to be implied. However I also know a definition that a function $f$ is positive definite if $sum_{ij} overline{z_i}z_j f(z_i-z_j)$ for any $z_iinBbb C$.
– s.harp
Sep 28 '17 at 9:03
add a comment |
What does it mean for a function to be positive (semi)definite?
Assume we have a positive (semi)definite function $K(X, Y)$ that lives in $R^n$, and maps vector pairs in $R^n$ into a scalar in R. For instance, a kernel function is defined as $K(x,y) = <phi(x),phi(y)>$ for some appropriate mapping function $phi$. Assume further that we have n vectors, $v_1, cdots, v_n$, with real entries. We say that a matrix, A, has the entries $A_{ij} = K(v_i,v_j) = A_{ji} = K(v_j,v_i)$. Why we can conclude that A is a positive (semi)definite matrix?
Why the sum of positive (semi)definite functions is also positive (semi)definite?
linear-algebra functional-analysis covariance positive-semidefinite
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
asked Oct 20 '15 at 20:40
Jack Shi
617
What does it mean for a function to be positive (semi)definite?
Assume we have a positive (semi)definite function $K(X, Y)$ that lives in $R^n$, and maps vector pairs in $R^n$ into a scalar in R. For instance, a kernel function is defined as $K(x,y) = <phi(x),phi(y)>$ for some appropriate mapping function $phi$. Assume further that we have n vectors, $v_1, cdots, v_n$, with real entries. We say that a matrix, A, has the entries $A_{ij} = K(v_i,v_j) = A_{ji} = K(v_j,v_i)$. Why we can conclude that A is a positive (semi)definite matrix?
Why the sum of positive (semi)definite functions is also positive (semi)definite?
linear-algebra functional-analysis covariance positive-semidefinite
linear-algebra functional-analysis covariance positive-semidefinite
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
asked Oct 20 '15 at 20:40
Jack Shi
617
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
asked Oct 20 '15 at 20:40
Jack Shi
617
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
edited Jul 15 at 6:02
Rodrigo de Azevedo
12.8k41855
12.8k41855
asked Oct 20 '15 at 20:40
Jack Shi
617
asked Oct 20 '15 at 20:40
Jack Shi
617
asked Oct 20 '15 at 20:40
Jack Shi
617
617
The question asks when a "function" is positive definite and then talks about bilinear forms, so I assume that to be implied. However I also know a definition that a function $f$ is positive definite if $sum_{ij} overline{z_i}z_j f(z_i-z_j)$ for any $z_iinBbb C$.
– s.harp
Sep 28 '17 at 9:03
add a comment |
The question asks when a "function" is positive definite and then talks about bilinear forms, so I assume that to be implied. However I also know a definition that a function $f$ is positive definite if $sum_{ij} overline{z_i}z_j f(z_i-z_j)$ for any $z_iinBbb C$.
– s.harp
Sep 28 '17 at 9:03
The question asks when a "function" is positive definite and then talks about bilinear forms, so I assume that to be implied. However I also know a definition that a function $f$ is positive definite if $sum_{ij} overline{z_i}z_j f(z_i-z_j)$ for any $z_iinBbb C$.
– s.harp
Sep 28 '17 at 9:03
The question asks when a "function" is positive definite and then talks about bilinear forms, so I assume that to be implied. However I also know a definition that a function $f$ is positive definite if $sum_{ij} overline{z_i}z_j f(z_i-z_j)$ for any $z_iinBbb C$.
– s.harp
Sep 28 '17 at 9:03
add a comment |
1 Answer
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votes
Recall that:
Definition. Let $X$ be a $defR{mathbf R}R$-vector space. A bilinear map $K colon X times X to R$ is called positive semi-definite, iff we have $K(x,x) ge 0$ for all $x in X$. If moreover $K(x,x) = 0 iff x= 0$, $K$ is called positive definite.
With that we have: Suppose, $K colon R^ntimes R^n to R$ is positive (semi)definite, let $v_1, ldots, v_n in R^n$ be $n$ vectors, then the matrix $A = bigl(K(v_i,v_j)bigr)_{i,j}$ is positive (semi-)definite, as for $xi in R^n$ we have, due to $K$'s bilinearity:
begin{align*}
def<#1>{left<#1right>}<Axi, xi> &= sum_{i=1}^n (Axi)_ixi_i\
&= sum_{i,j=1}^n A_{ij}xi_jxi_i\
&= sum_{i,j=1}^n xi_jxi_i K(v_i, v_j)\
&= Kleft(sum_i xi_i v_i, sum_j xi_j v_jright)\
&ge 0
end{align*}
If $K$ is positive definite and the $v_i$'s are linear independent, then $A$ is positive definite: Suppose $<Axi, xi> = 0$, then by the above, we have $K(sum_i xi_i v_i, sum_i xi_i v_i) = 0$, hence - as $K$ is definite - $sum_i xi_i v_i = 0$. As the $v_i$ are independent, this implies $xi = 0$. So $A$ is positive definite.
answered Oct 21 '15 at 7:46
martini
70.3k45990
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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votes
Recall that:
Definition. Let $X$ be a $defR{mathbf R}R$-vector space. A bilinear map $K colon X times X to R$ is called positive semi-definite, iff we have $K(x,x) ge 0$ for all $x in X$. If moreover $K(x,x) = 0 iff x= 0$, $K$ is called positive definite.
With that we have: Suppose, $K colon R^ntimes R^n to R$ is positive (semi)definite, let $v_1, ldots, v_n in R^n$ be $n$ vectors, then the matrix $A = bigl(K(v_i,v_j)bigr)_{i,j}$ is positive (semi-)definite, as for $xi in R^n$ we have, due to $K$'s bilinearity:
begin{align*}
def<#1>{left<#1right>}<Axi, xi> &= sum_{i=1}^n (Axi)_ixi_i\
&= sum_{i,j=1}^n A_{ij}xi_jxi_i\
&= sum_{i,j=1}^n xi_jxi_i K(v_i, v_j)\
&= Kleft(sum_i xi_i v_i, sum_j xi_j v_jright)\
&ge 0
end{align*}
If $K$ is positive definite and the $v_i$'s are linear independent, then $A$ is positive definite: Suppose $<Axi, xi> = 0$, then by the above, we have $K(sum_i xi_i v_i, sum_i xi_i v_i) = 0$, hence - as $K$ is definite - $sum_i xi_i v_i = 0$. As the $v_i$ are independent, this implies $xi = 0$. So $A$ is positive definite.
answered Oct 21 '15 at 7:46
martini
70.3k45990
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
add a comment |
Recall that:
Definition. Let $X$ be a $defR{mathbf R}R$-vector space. A bilinear map $K colon X times X to R$ is called positive semi-definite, iff we have $K(x,x) ge 0$ for all $x in X$. If moreover $K(x,x) = 0 iff x= 0$, $K$ is called positive definite.
With that we have: Suppose, $K colon R^ntimes R^n to R$ is positive (semi)definite, let $v_1, ldots, v_n in R^n$ be $n$ vectors, then the matrix $A = bigl(K(v_i,v_j)bigr)_{i,j}$ is positive (semi-)definite, as for $xi in R^n$ we have, due to $K$'s bilinearity:
begin{align*}
def<#1>{left<#1right>}<Axi, xi> &= sum_{i=1}^n (Axi)_ixi_i\
&= sum_{i,j=1}^n A_{ij}xi_jxi_i\
&= sum_{i,j=1}^n xi_jxi_i K(v_i, v_j)\
&= Kleft(sum_i xi_i v_i, sum_j xi_j v_jright)\
&ge 0
end{align*}
If $K$ is positive definite and the $v_i$'s are linear independent, then $A$ is positive definite: Suppose $<Axi, xi> = 0$, then by the above, we have $K(sum_i xi_i v_i, sum_i xi_i v_i) = 0$, hence - as $K$ is definite - $sum_i xi_i v_i = 0$. As the $v_i$ are independent, this implies $xi = 0$. So $A$ is positive definite.
answered Oct 21 '15 at 7:46
martini
70.3k45990
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
add a comment |
Recall that:
Definition. Let $X$ be a $defR{mathbf R}R$-vector space. A bilinear map $K colon X times X to R$ is called positive semi-definite, iff we have $K(x,x) ge 0$ for all $x in X$. If moreover $K(x,x) = 0 iff x= 0$, $K$ is called positive definite.
With that we have: Suppose, $K colon R^ntimes R^n to R$ is positive (semi)definite, let $v_1, ldots, v_n in R^n$ be $n$ vectors, then the matrix $A = bigl(K(v_i,v_j)bigr)_{i,j}$ is positive (semi-)definite, as for $xi in R^n$ we have, due to $K$'s bilinearity:
begin{align*}
def<#1>{left<#1right>}<Axi, xi> &= sum_{i=1}^n (Axi)_ixi_i\
&= sum_{i,j=1}^n A_{ij}xi_jxi_i\
&= sum_{i,j=1}^n xi_jxi_i K(v_i, v_j)\
&= Kleft(sum_i xi_i v_i, sum_j xi_j v_jright)\
&ge 0
end{align*}
If $K$ is positive definite and the $v_i$'s are linear independent, then $A$ is positive definite: Suppose $<Axi, xi> = 0$, then by the above, we have $K(sum_i xi_i v_i, sum_i xi_i v_i) = 0$, hence - as $K$ is definite - $sum_i xi_i v_i = 0$. As the $v_i$ are independent, this implies $xi = 0$. So $A$ is positive definite.
answered Oct 21 '15 at 7:46
martini
70.3k45990
Recall that:
Definition. Let $X$ be a $defR{mathbf R}R$-vector space. A bilinear map $K colon X times X to R$ is called positive semi-definite, iff we have $K(x,x) ge 0$ for all $x in X$. If moreover $K(x,x) = 0 iff x= 0$, $K$ is called positive definite.
With that we have: Suppose, $K colon R^ntimes R^n to R$ is positive (semi)definite, let $v_1, ldots, v_n in R^n$ be $n$ vectors, then the matrix $A = bigl(K(v_i,v_j)bigr)_{i,j}$ is positive (semi-)definite, as for $xi in R^n$ we have, due to $K$'s bilinearity:
begin{align*}
def<#1>{left<#1right>}<Axi, xi> &= sum_{i=1}^n (Axi)_ixi_i\
&= sum_{i,j=1}^n A_{ij}xi_jxi_i\
&= sum_{i,j=1}^n xi_jxi_i K(v_i, v_j)\
&= Kleft(sum_i xi_i v_i, sum_j xi_j v_jright)\
&ge 0
end{align*}
If $K$ is positive definite and the $v_i$'s are linear independent, then $A$ is positive definite: Suppose $<Axi, xi> = 0$, then by the above, we have $K(sum_i xi_i v_i, sum_i xi_i v_i) = 0$, hence - as $K$ is definite - $sum_i xi_i v_i = 0$. As the $v_i$ are independent, this implies $xi = 0$. So $A$ is positive definite.
answered Oct 21 '15 at 7:46
martini
70.3k45990
answered Oct 21 '15 at 7:46
martini
70.3k45990
answered Oct 21 '15 at 7:46
martini
70.3k45990
answered Oct 21 '15 at 7:46
martini
70.3k45990
70.3k45990
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
add a comment |
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
One question: Where did you find the definition above?
– Jack Shi
Oct 21 '15 at 7:57
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
It is a well established definition, what do you mean by "find"?
– martini
Oct 21 '15 at 8:00
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
I'm not familiar with positive definite functions, and when I searched online, I didn't find any information.
– Jack Shi
Oct 21 '15 at 8:02
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
en.wikipedia.org/wiki/Definite_quadratic_form
– martini
Oct 21 '15 at 8:16
add a comment |
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The question asks when a "function" is positive definite and then talks about bilinear forms, so I assume that to be implied. However I also know a definition that a function $f$ is positive definite if $sum_{ij} overline{z_i}z_j f(z_i-z_j)$ for any $z_iinBbb C$.
– s.harp
Sep 28 '17 at 9:03