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Equivalent notation for a strictly convex function is not strictly convex
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Let $x in mathbbR^n$
and, we define the following function: $f(x) = sum_i=1^n (n-1-0.1i)(x_i)^2$.
This function is a convex function since it can be written as $x^T Q x $ where
$Q = beginpmatrix n -1-0.1i & 0 & ldots & 0 \ 0 & n-1 - 0.1i & ldots & 0 \ vdots & vdots & ddots& vdots \ 0 & ldots & & n-1-0.1i endpmatrix \$
is a positive definite matrix.
I am trying to reformulate this function by using $x in mathbbR^n_+$ only. Thus, I rewrite $x_i = x_i^+ - x_i ^ -$ where $x_i^+,x_i^-in mathbbR_+$
So I write $f(x) = sum_i=1^n (n-1-0.1i)(x_i^+ - x_i^-)^2$ and $Q$ becomes:
$Q^* = beginpmatrix n-1-0.1 i & -(n-1-0.1i) & 0 & 0 & ldots \
-(n-1 -0.1i) & n-1-0.1i & 0 & 0 & ldots \
0 & 0 & n-1-0.1i & -(n-1-0.1i) & ldots \
0 & 0 & -(n-1-0.1i) & n-1-0.1i & ldots \
vdots & vdots & vdots & vdots & vdots endpmatrix$
The new $Q^*$ is not positive definite, so $f(x)$ is not convex. How can I get this equivalent notation as strictly convex?
convex-analysis positive-semidefinite
edited Jul 27 at 6:40
max_zorn
3,15151028
asked Jul 23 at 23:02
aslv95
397
 |Â
show 2 more comments
up vote
0
down vote
favorite
Let $x in mathbbR^n$
and, we define the following function: $f(x) = sum_i=1^n (n-1-0.1i)(x_i)^2$.
This function is a convex function since it can be written as $x^T Q x $ where
$Q = beginpmatrix n -1-0.1i & 0 & ldots & 0 \ 0 & n-1 - 0.1i & ldots & 0 \ vdots & vdots & ddots& vdots \ 0 & ldots & & n-1-0.1i endpmatrix \$
is a positive definite matrix.
I am trying to reformulate this function by using $x in mathbbR^n_+$ only. Thus, I rewrite $x_i = x_i^+ - x_i ^ -$ where $x_i^+,x_i^-in mathbbR_+$
So I write $f(x) = sum_i=1^n (n-1-0.1i)(x_i^+ - x_i^-)^2$ and $Q$ becomes:
$Q^* = beginpmatrix n-1-0.1 i & -(n-1-0.1i) & 0 & 0 & ldots \
-(n-1 -0.1i) & n-1-0.1i & 0 & 0 & ldots \
0 & 0 & n-1-0.1i & -(n-1-0.1i) & ldots \
0 & 0 & -(n-1-0.1i) & n-1-0.1i & ldots \
vdots & vdots & vdots & vdots & vdots endpmatrix$
The new $Q^*$ is not positive definite, so $f(x)$ is not convex. How can I get this equivalent notation as strictly convex?
convex-analysis positive-semidefinite
edited Jul 27 at 6:40
max_zorn
3,15151028
asked Jul 23 at 23:02
aslv95
397
It looks positive semidefinite to me.
– Rahul
Jul 24 at 4:49
In Matlab I am trying to get the Cholesky Decomp and it says: Error using chol Matrix must be positive definite.
– aslv95
Jul 24 at 12:56
Cholesky only works on positive definite matrices, you have a positive semi-definite matrix.
– Johan Löfberg
Jul 24 at 13:40
Eventually, I want to use the Chol Decomp. Does this imply equivalent notations can lose positive definiteness?
– aslv95
Jul 24 at 15:49
$Q$ should not contain $i$ btw
– LinAlg
Jul 24 at 16:19
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $x in mathbbR^n$
and, we define the following function: $f(x) = sum_i=1^n (n-1-0.1i)(x_i)^2$.
This function is a convex function since it can be written as $x^T Q x $ where
$Q = beginpmatrix n -1-0.1i & 0 & ldots & 0 \ 0 & n-1 - 0.1i & ldots & 0 \ vdots & vdots & ddots& vdots \ 0 & ldots & & n-1-0.1i endpmatrix \$
is a positive definite matrix.
I am trying to reformulate this function by using $x in mathbbR^n_+$ only. Thus, I rewrite $x_i = x_i^+ - x_i ^ -$ where $x_i^+,x_i^-in mathbbR_+$
So I write $f(x) = sum_i=1^n (n-1-0.1i)(x_i^+ - x_i^-)^2$ and $Q$ becomes:
$Q^* = beginpmatrix n-1-0.1 i & -(n-1-0.1i) & 0 & 0 & ldots \
-(n-1 -0.1i) & n-1-0.1i & 0 & 0 & ldots \
0 & 0 & n-1-0.1i & -(n-1-0.1i) & ldots \
0 & 0 & -(n-1-0.1i) & n-1-0.1i & ldots \
vdots & vdots & vdots & vdots & vdots endpmatrix$
The new $Q^*$ is not positive definite, so $f(x)$ is not convex. How can I get this equivalent notation as strictly convex?
convex-analysis positive-semidefinite
edited Jul 27 at 6:40
max_zorn
3,15151028
asked Jul 23 at 23:02
aslv95
397
Let $x in mathbbR^n$
and, we define the following function: $f(x) = sum_i=1^n (n-1-0.1i)(x_i)^2$.
This function is a convex function since it can be written as $x^T Q x $ where
$Q = beginpmatrix n -1-0.1i & 0 & ldots & 0 \ 0 & n-1 - 0.1i & ldots & 0 \ vdots & vdots & ddots& vdots \ 0 & ldots & & n-1-0.1i endpmatrix \$
is a positive definite matrix.
I am trying to reformulate this function by using $x in mathbbR^n_+$ only. Thus, I rewrite $x_i = x_i^+ - x_i ^ -$ where $x_i^+,x_i^-in mathbbR_+$
So I write $f(x) = sum_i=1^n (n-1-0.1i)(x_i^+ - x_i^-)^2$ and $Q$ becomes:
$Q^* = beginpmatrix n-1-0.1 i & -(n-1-0.1i) & 0 & 0 & ldots \
-(n-1 -0.1i) & n-1-0.1i & 0 & 0 & ldots \
0 & 0 & n-1-0.1i & -(n-1-0.1i) & ldots \
0 & 0 & -(n-1-0.1i) & n-1-0.1i & ldots \
vdots & vdots & vdots & vdots & vdots endpmatrix$
The new $Q^*$ is not positive definite, so $f(x)$ is not convex. How can I get this equivalent notation as strictly convex?
convex-analysis positive-semidefinite
edited Jul 27 at 6:40
max_zorn
3,15151028
asked Jul 23 at 23:02
aslv95
397
edited Jul 27 at 6:40
max_zorn
3,15151028
edited Jul 27 at 6:40
max_zorn
3,15151028
edited Jul 27 at 6:40
max_zorn
3,15151028
3,15151028
asked Jul 23 at 23:02
aslv95
397
asked Jul 23 at 23:02
aslv95
397
asked Jul 23 at 23:02
aslv95
397
397
It looks positive semidefinite to me.
– Rahul
Jul 24 at 4:49
In Matlab I am trying to get the Cholesky Decomp and it says: Error using chol Matrix must be positive definite.
– aslv95
Jul 24 at 12:56
Cholesky only works on positive definite matrices, you have a positive semi-definite matrix.
– Johan Löfberg
Jul 24 at 13:40
Eventually, I want to use the Chol Decomp. Does this imply equivalent notations can lose positive definiteness?
– aslv95
Jul 24 at 15:49
$Q$ should not contain $i$ btw
– LinAlg
Jul 24 at 16:19
 |Â
show 2 more comments
It looks positive semidefinite to me.
– Rahul
Jul 24 at 4:49
In Matlab I am trying to get the Cholesky Decomp and it says: Error using chol Matrix must be positive definite.
– aslv95
Jul 24 at 12:56
Cholesky only works on positive definite matrices, you have a positive semi-definite matrix.
– Johan Löfberg
Jul 24 at 13:40
Eventually, I want to use the Chol Decomp. Does this imply equivalent notations can lose positive definiteness?
– aslv95
Jul 24 at 15:49
$Q$ should not contain $i$ btw
– LinAlg
Jul 24 at 16:19
It looks positive semidefinite to me.
– Rahul
Jul 24 at 4:49
It looks positive semidefinite to me.
– Rahul
Jul 24 at 4:49
In Matlab I am trying to get the Cholesky Decomp and it says: Error using chol Matrix must be positive definite.
– aslv95
Jul 24 at 12:56
In Matlab I am trying to get the Cholesky Decomp and it says: Error using chol Matrix must be positive definite.
– aslv95
Jul 24 at 12:56
Cholesky only works on positive definite matrices, you have a positive semi-definite matrix.
– Johan Löfberg
Jul 24 at 13:40
Cholesky only works on positive definite matrices, you have a positive semi-definite matrix.
– Johan Löfberg
Jul 24 at 13:40
Eventually, I want to use the Chol Decomp. Does this imply equivalent notations can lose positive definiteness?
– aslv95
Jul 24 at 15:49
Eventually, I want to use the Chol Decomp. Does this imply equivalent notations can lose positive definiteness?
– aslv95
Jul 24 at 15:49
$Q$ should not contain $i$ btw
– LinAlg
Jul 24 at 16:19
$Q$ should not contain $i$ btw
– LinAlg
Jul 24 at 16:19
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your new function is not strictly convex, since it is constant on the line $x_i^+ = x_i^-$. However, your function is still convex, as a convex function of an affine function is convex (easily proved with the definition of convexity).
answered Jul 24 at 16:20
LinAlg
5,4111319
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
Your new function is not strictly convex, since it is constant on the line $x_i^+ = x_i^-$. However, your function is still convex, as a convex function of an affine function is convex (easily proved with the definition of convexity).
answered Jul 24 at 16:20
LinAlg
5,4111319
add a comment |Â
up vote
1
down vote
accepted
Your new function is not strictly convex, since it is constant on the line $x_i^+ = x_i^-$. However, your function is still convex, as a convex function of an affine function is convex (easily proved with the definition of convexity).
answered Jul 24 at 16:20
LinAlg
5,4111319
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your new function is not strictly convex, since it is constant on the line $x_i^+ = x_i^-$. However, your function is still convex, as a convex function of an affine function is convex (easily proved with the definition of convexity).
answered Jul 24 at 16:20
LinAlg
5,4111319
Your new function is not strictly convex, since it is constant on the line $x_i^+ = x_i^-$. However, your function is still convex, as a convex function of an affine function is convex (easily proved with the definition of convexity).
answered Jul 24 at 16:20
LinAlg
5,4111319
answered Jul 24 at 16:20
LinAlg
5,4111319
answered Jul 24 at 16:20
LinAlg
5,4111319
answered Jul 24 at 16:20
LinAlg
5,4111319
5,4111319
add a comment |Â
add a comment |Â
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It looks positive semidefinite to me.
– Rahul
Jul 24 at 4:49
In Matlab I am trying to get the Cholesky Decomp and it says: Error using chol Matrix must be positive definite.
– aslv95
Jul 24 at 12:56
Cholesky only works on positive definite matrices, you have a positive semi-definite matrix.
– Johan Löfberg
Jul 24 at 13:40
Eventually, I want to use the Chol Decomp. Does this imply equivalent notations can lose positive definiteness?
– aslv95
Jul 24 at 15:49
$Q$ should not contain $i$ btw
– LinAlg
Jul 24 at 16:19