Why is $|A-BCA|_F^2$ non-convex regarding $B,C$?
$begingroup$
I noticed somewhere, a function similar to $|A-BCA|_F^2$ was claimed to be non-convex regarding $B$ or $C$ individually, while a relaxation like $D=CA$ makes $|A-BD|_F^2$ convex respect to each $B$ and $D$ individually. The matrices $A, B, C, D$ are not square-matrix, and by individually, I mean assuming the value of one matrix fixed while analyzing w.r.t the other matrix.
Honestly, I do not understand why the first one is non-convex while the latter is convex?
Update:
In fact, i fund the above claim in
This NIPS paper, where they claimed that Eq. 5 is not a convex problem w.r.t $P_k$, but the relaxation in Eq. 6 makes the objective convex w.r.t to each of the components.
linear-algebra derivatives convex-analysis
asked Jan 22 at 23:33
BabakBabak
354111
$endgroup$
add a comment |
$begingroup$
I noticed somewhere, a function similar to $|A-BCA|_F^2$ was claimed to be non-convex regarding $B$ or $C$ individually, while a relaxation like $D=CA$ makes $|A-BD|_F^2$ convex respect to each $B$ and $D$ individually. The matrices $A, B, C, D$ are not square-matrix, and by individually, I mean assuming the value of one matrix fixed while analyzing w.r.t the other matrix.
Honestly, I do not understand why the first one is non-convex while the latter is convex?
Update:
In fact, i fund the above claim in
This NIPS paper, where they claimed that Eq. 5 is not a convex problem w.r.t $P_k$, but the relaxation in Eq. 6 makes the objective convex w.r.t to each of the components.
linear-algebra derivatives convex-analysis
asked Jan 22 at 23:33
BabakBabak
354111
$endgroup$
$begingroup$
I doubt that the latter function is convex. Even in the one-dimensional case, $f(x,y) = (1-xy)^2$ is not convex in $(x,y)$.
$endgroup$
– gerw
Jan 23 at 10:43
$begingroup$
@gerw: I meant considering convexity w.r.t one matrix at a time. In your 1-D example, $f(x,y)$ is convex w.r.t $x$ or $y$ individually.
$endgroup$
– Babak
Jan 27 at 17:41
$begingroup$
If you are interested in this separate convexity, also the first function should be convex (w.r.t. this notion).
$endgroup$
– gerw
Jan 27 at 19:49
$begingroup$
@gerw: how can you show that?!
$endgroup$
– Babak
Jan 27 at 20:27
$begingroup$
@gerw: I'm especially interested in the convexity of the first function regarding $C$!
$endgroup$
– Babak
Jan 27 at 20:34
add a comment |
$begingroup$
I noticed somewhere, a function similar to $|A-BCA|_F^2$ was claimed to be non-convex regarding $B$ or $C$ individually, while a relaxation like $D=CA$ makes $|A-BD|_F^2$ convex respect to each $B$ and $D$ individually. The matrices $A, B, C, D$ are not square-matrix, and by individually, I mean assuming the value of one matrix fixed while analyzing w.r.t the other matrix.
Honestly, I do not understand why the first one is non-convex while the latter is convex?
Update:
In fact, i fund the above claim in
This NIPS paper, where they claimed that Eq. 5 is not a convex problem w.r.t $P_k$, but the relaxation in Eq. 6 makes the objective convex w.r.t to each of the components.
linear-algebra derivatives convex-analysis
asked Jan 22 at 23:33
BabakBabak
354111
$endgroup$
I noticed somewhere, a function similar to $|A-BCA|_F^2$ was claimed to be non-convex regarding $B$ or $C$ individually, while a relaxation like $D=CA$ makes $|A-BD|_F^2$ convex respect to each $B$ and $D$ individually. The matrices $A, B, C, D$ are not square-matrix, and by individually, I mean assuming the value of one matrix fixed while analyzing w.r.t the other matrix.
Honestly, I do not understand why the first one is non-convex while the latter is convex?
Update:
In fact, i fund the above claim in
This NIPS paper, where they claimed that Eq. 5 is not a convex problem w.r.t $P_k$, but the relaxation in Eq. 6 makes the objective convex w.r.t to each of the components.
linear-algebra derivatives convex-analysis
linear-algebra derivatives convex-analysis
asked Jan 22 at 23:33
BabakBabak
354111
asked Jan 22 at 23:33
BabakBabak
354111
edited Jan 28 at 9:54
Babak
asked Jan 22 at 23:33
BabakBabak
354111
asked Jan 22 at 23:33
BabakBabak
354111
asked Jan 22 at 23:33
BabakBabak
354111
354111
$begingroup$
I doubt that the latter function is convex. Even in the one-dimensional case, $f(x,y) = (1-xy)^2$ is not convex in $(x,y)$.
$endgroup$
– gerw
Jan 23 at 10:43
$begingroup$
@gerw: I meant considering convexity w.r.t one matrix at a time. In your 1-D example, $f(x,y)$ is convex w.r.t $x$ or $y$ individually.
$endgroup$
– Babak
Jan 27 at 17:41
$begingroup$
If you are interested in this separate convexity, also the first function should be convex (w.r.t. this notion).
$endgroup$
– gerw
Jan 27 at 19:49
$begingroup$
@gerw: how can you show that?!
$endgroup$
– Babak
Jan 27 at 20:27
$begingroup$
@gerw: I'm especially interested in the convexity of the first function regarding $C$!
$endgroup$
– Babak
Jan 27 at 20:34
add a comment |
$begingroup$
I doubt that the latter function is convex. Even in the one-dimensional case, $f(x,y) = (1-xy)^2$ is not convex in $(x,y)$.
$endgroup$
– gerw
Jan 23 at 10:43
$begingroup$
@gerw: I meant considering convexity w.r.t one matrix at a time. In your 1-D example, $f(x,y)$ is convex w.r.t $x$ or $y$ individually.
$endgroup$
– Babak
Jan 27 at 17:41
$begingroup$
If you are interested in this separate convexity, also the first function should be convex (w.r.t. this notion).
$endgroup$
– gerw
Jan 27 at 19:49
$begingroup$
@gerw: how can you show that?!
$endgroup$
– Babak
Jan 27 at 20:27
$begingroup$
@gerw: I'm especially interested in the convexity of the first function regarding $C$!
$endgroup$
– Babak
Jan 27 at 20:34
$begingroup$
I doubt that the latter function is convex. Even in the one-dimensional case, $f(x,y) = (1-xy)^2$ is not convex in $(x,y)$.
$endgroup$
– gerw
Jan 23 at 10:43
$begingroup$
I doubt that the latter function is convex. Even in the one-dimensional case, $f(x,y) = (1-xy)^2$ is not convex in $(x,y)$.
$endgroup$
– gerw
Jan 23 at 10:43
$begingroup$
@gerw: I meant considering convexity w.r.t one matrix at a time. In your 1-D example, $f(x,y)$ is convex w.r.t $x$ or $y$ individually.
$endgroup$
– Babak
Jan 27 at 17:41
$begingroup$
@gerw: I meant considering convexity w.r.t one matrix at a time. In your 1-D example, $f(x,y)$ is convex w.r.t $x$ or $y$ individually.
$endgroup$
– Babak
Jan 27 at 17:41
$begingroup$
If you are interested in this separate convexity, also the first function should be convex (w.r.t. this notion).
$endgroup$
– gerw
Jan 27 at 19:49
$begingroup$
If you are interested in this separate convexity, also the first function should be convex (w.r.t. this notion).
$endgroup$
– gerw
Jan 27 at 19:49
$begingroup$
@gerw: how can you show that?!
$endgroup$
– Babak
Jan 27 at 20:27
$begingroup$
@gerw: how can you show that?!
$endgroup$
– Babak
Jan 27 at 20:27
$begingroup$
@gerw: I'm especially interested in the convexity of the first function regarding $C$!
$endgroup$
– Babak
Jan 27 at 20:34
$begingroup$
@gerw: I'm especially interested in the convexity of the first function regarding $C$!
$endgroup$
– Babak
Jan 27 at 20:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us check that the first function is convex w.r.t. $C$. Thus, we have to show that the function
$$t mapsto | A - B , (C + t , D) , A|_F$$
is convex for all matrices $D$. Let us abbreviate $G := A - B , C , A$ and $H := B , D , A$. Then,
$$
|A - B , (C + t , D) , A|_F
=
|G - t , H|_F^2
=
|G|_F^2 - 2 , t , (G,H)_F + t^2 , |H|_F^2$$
and this function is clearly convex w.r.t. $t$. Here, $(cdot,cdot)_F$ is the Frobenius inner product.
answered Jan 28 at 7:13
gerwgerw
19.6k11334
$endgroup$
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Let us check that the first function is convex w.r.t. $C$. Thus, we have to show that the function
$$t mapsto | A - B , (C + t , D) , A|_F$$
is convex for all matrices $D$. Let us abbreviate $G := A - B , C , A$ and $H := B , D , A$. Then,
$$
|A - B , (C + t , D) , A|_F
=
|G - t , H|_F^2
=
|G|_F^2 - 2 , t , (G,H)_F + t^2 , |H|_F^2$$
and this function is clearly convex w.r.t. $t$. Here, $(cdot,cdot)_F$ is the Frobenius inner product.
answered Jan 28 at 7:13
gerwgerw
19.6k11334
$endgroup$
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
add a comment |
$begingroup$
Let us check that the first function is convex w.r.t. $C$. Thus, we have to show that the function
$$t mapsto | A - B , (C + t , D) , A|_F$$
is convex for all matrices $D$. Let us abbreviate $G := A - B , C , A$ and $H := B , D , A$. Then,
$$
|A - B , (C + t , D) , A|_F
=
|G - t , H|_F^2
=
|G|_F^2 - 2 , t , (G,H)_F + t^2 , |H|_F^2$$
and this function is clearly convex w.r.t. $t$. Here, $(cdot,cdot)_F$ is the Frobenius inner product.
answered Jan 28 at 7:13
gerwgerw
19.6k11334
$endgroup$
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
add a comment |
$begingroup$
Let us check that the first function is convex w.r.t. $C$. Thus, we have to show that the function
$$t mapsto | A - B , (C + t , D) , A|_F$$
is convex for all matrices $D$. Let us abbreviate $G := A - B , C , A$ and $H := B , D , A$. Then,
$$
|A - B , (C + t , D) , A|_F
=
|G - t , H|_F^2
=
|G|_F^2 - 2 , t , (G,H)_F + t^2 , |H|_F^2$$
and this function is clearly convex w.r.t. $t$. Here, $(cdot,cdot)_F$ is the Frobenius inner product.
answered Jan 28 at 7:13
gerwgerw
19.6k11334
$endgroup$
Let us check that the first function is convex w.r.t. $C$. Thus, we have to show that the function
$$t mapsto | A - B , (C + t , D) , A|_F$$
is convex for all matrices $D$. Let us abbreviate $G := A - B , C , A$ and $H := B , D , A$. Then,
$$
|A - B , (C + t , D) , A|_F
=
|G - t , H|_F^2
=
|G|_F^2 - 2 , t , (G,H)_F + t^2 , |H|_F^2$$
and this function is clearly convex w.r.t. $t$. Here, $(cdot,cdot)_F$ is the Frobenius inner product.
answered Jan 28 at 7:13
gerwgerw
19.6k11334
answered Jan 28 at 7:13
gerwgerw
19.6k11334
answered Jan 28 at 7:13
gerwgerw
19.6k11334
answered Jan 28 at 7:13
gerwgerw
19.6k11334
19.6k11334
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
add a comment |
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
So, you are saying that a function $f(X)$ would be convex w.r.t matrix $X$ if $f(X+tY)$ is convex w.r.t the scalar $t$?
$endgroup$
– Babak
Jan 28 at 9:50
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
Also, does it mean that they made a wrong claim in papers.nips.cc/paper/… regarding the convexity of Eq. 5 w.r.t each of its component?
$endgroup$
– Babak
Jan 28 at 9:59
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
I do not see any statement w.r.t. separate convexity of (5). They only say that (5) is not convex.
$endgroup$
– gerw
Jan 28 at 10:44
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
$begingroup$
Then what was the point of converting (5) to (6) given (5) is separately convex? Couldn't they apply the same optimization procedure directing on (5)?
$endgroup$
– Babak
Jan 28 at 11:07
add a comment |
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$begingroup$
I doubt that the latter function is convex. Even in the one-dimensional case, $f(x,y) = (1-xy)^2$ is not convex in $(x,y)$.
$endgroup$
– gerw
Jan 23 at 10:43
$begingroup$
@gerw: I meant considering convexity w.r.t one matrix at a time. In your 1-D example, $f(x,y)$ is convex w.r.t $x$ or $y$ individually.
$endgroup$
– Babak
Jan 27 at 17:41
$begingroup$
If you are interested in this separate convexity, also the first function should be convex (w.r.t. this notion).
$endgroup$
– gerw
Jan 27 at 19:49
$begingroup$
@gerw: how can you show that?!
$endgroup$
– Babak
Jan 27 at 20:27
$begingroup$
@gerw: I'm especially interested in the convexity of the first function regarding $C$!
$endgroup$
– Babak
Jan 27 at 20:34