How to prove a matrix function is convex or nonconvex?
$begingroup$
I have a function of three matrix variables. But now, the authors fix two of them, then update one, and I cannot understand how this function is convex in each iteration in the paper.
This formula is :
$$f(W,V,B) =|XW-V|^2_F +|Y-VB|^2_F +operatorname{tr}(V'LV) +2operatorname{tr}(W'DW),$$
where $X$, $Y$ are constant matrices and $L$ is constant laplace matrix. Suppose $D$ is a constant diagonal matrix.
Now, we fix two variables $W$ and $V$, then update $B$. How to solve?
If we do not fix any variables, how to explain that the objective function is non-convex?
In general, we use the Hessian matrix , but what should I do when the variable is a matrix?
convex-analysis convex-optimization machine-learning non-convex-optimization
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
$endgroup$
add a comment |
$begingroup$
I have a function of three matrix variables. But now, the authors fix two of them, then update one, and I cannot understand how this function is convex in each iteration in the paper.
This formula is :
$$f(W,V,B) =|XW-V|^2_F +|Y-VB|^2_F +operatorname{tr}(V'LV) +2operatorname{tr}(W'DW),$$
where $X$, $Y$ are constant matrices and $L$ is constant laplace matrix. Suppose $D$ is a constant diagonal matrix.
Now, we fix two variables $W$ and $V$, then update $B$. How to solve?
If we do not fix any variables, how to explain that the objective function is non-convex?
In general, we use the Hessian matrix , but what should I do when the variable is a matrix?
convex-analysis convex-optimization machine-learning non-convex-optimization
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 7:17
$begingroup$
B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 9:29
$begingroup$
@Michal Adamaszek If the function is non-convex, we can use the anti-evidence method. But when only B is a matrix variable, the objective function is a convex function. Is it appropriate to use this anti-evidence method: (y-vb)^ 2?
$endgroup$
– learn_truth
Nov 28 '18 at 12:44
$begingroup$
@lilili My hint was not about proving general non-convexity. It was about how to prove convexity in B: Frobenius norm square is convex and the rest is linear.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 13:24
$begingroup$
@Michal Adamaszek The variable is a matrix, does the rule still apply?
$endgroup$
– learn_truth
Nov 29 '18 at 3:09
add a comment |
$begingroup$
I have a function of three matrix variables. But now, the authors fix two of them, then update one, and I cannot understand how this function is convex in each iteration in the paper.
This formula is :
$$f(W,V,B) =|XW-V|^2_F +|Y-VB|^2_F +operatorname{tr}(V'LV) +2operatorname{tr}(W'DW),$$
where $X$, $Y$ are constant matrices and $L$ is constant laplace matrix. Suppose $D$ is a constant diagonal matrix.
Now, we fix two variables $W$ and $V$, then update $B$. How to solve?
If we do not fix any variables, how to explain that the objective function is non-convex?
In general, we use the Hessian matrix , but what should I do when the variable is a matrix?
convex-analysis convex-optimization machine-learning non-convex-optimization
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
$endgroup$
I have a function of three matrix variables. But now, the authors fix two of them, then update one, and I cannot understand how this function is convex in each iteration in the paper.
This formula is :
$$f(W,V,B) =|XW-V|^2_F +|Y-VB|^2_F +operatorname{tr}(V'LV) +2operatorname{tr}(W'DW),$$
where $X$, $Y$ are constant matrices and $L$ is constant laplace matrix. Suppose $D$ is a constant diagonal matrix.
Now, we fix two variables $W$ and $V$, then update $B$. How to solve?
If we do not fix any variables, how to explain that the objective function is non-convex?
In general, we use the Hessian matrix , but what should I do when the variable is a matrix?
convex-analysis convex-optimization machine-learning non-convex-optimization
convex-analysis convex-optimization machine-learning non-convex-optimization
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
edited Dec 9 '18 at 12:57
learn_truth
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
asked Nov 28 '18 at 7:11
learn_truthlearn_truth
12
12
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 7:17
$begingroup$
B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 9:29
$begingroup$
@Michal Adamaszek If the function is non-convex, we can use the anti-evidence method. But when only B is a matrix variable, the objective function is a convex function. Is it appropriate to use this anti-evidence method: (y-vb)^ 2?
$endgroup$
– learn_truth
Nov 28 '18 at 12:44
$begingroup$
@lilili My hint was not about proving general non-convexity. It was about how to prove convexity in B: Frobenius norm square is convex and the rest is linear.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 13:24
$begingroup$
@Michal Adamaszek The variable is a matrix, does the rule still apply?
$endgroup$
– learn_truth
Nov 29 '18 at 3:09
add a comment |
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 7:17
$begingroup$
B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 9:29
$begingroup$
@Michal Adamaszek If the function is non-convex, we can use the anti-evidence method. But when only B is a matrix variable, the objective function is a convex function. Is it appropriate to use this anti-evidence method: (y-vb)^ 2?
$endgroup$
– learn_truth
Nov 28 '18 at 12:44
$begingroup$
@lilili My hint was not about proving general non-convexity. It was about how to prove convexity in B: Frobenius norm square is convex and the rest is linear.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 13:24
$begingroup$
@Michal Adamaszek The variable is a matrix, does the rule still apply?
$endgroup$
– learn_truth
Nov 29 '18 at 3:09
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 7:17
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 7:17
$begingroup$
B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 9:29
$begingroup$
B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 9:29
$begingroup$
@Michal Adamaszek If the function is non-convex, we can use the anti-evidence method. But when only B is a matrix variable, the objective function is a convex function. Is it appropriate to use this anti-evidence method: (y-vb)^ 2?
$endgroup$
– learn_truth
Nov 28 '18 at 12:44
$begingroup$
@Michal Adamaszek If the function is non-convex, we can use the anti-evidence method. But when only B is a matrix variable, the objective function is a convex function. Is it appropriate to use this anti-evidence method: (y-vb)^ 2?
$endgroup$
– learn_truth
Nov 28 '18 at 12:44
$begingroup$
@lilili My hint was not about proving general non-convexity. It was about how to prove convexity in B: Frobenius norm square is convex and the rest is linear.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 13:24
$begingroup$
@lilili My hint was not about proving general non-convexity. It was about how to prove convexity in B: Frobenius norm square is convex and the rest is linear.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 13:24
$begingroup$
@Michal Adamaszek The variable is a matrix, does the rule still apply?
$endgroup$
– learn_truth
Nov 29 '18 at 3:09
$begingroup$
@Michal Adamaszek The variable is a matrix, does the rule still apply?
$endgroup$
– learn_truth
Nov 29 '18 at 3:09
add a comment |
1 Answer
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$begingroup$
I want to give an answer to the first question, expanding what was already hinted. The goal is to prove that the function $B mapsto f(V,W,B) $ is convex for every $V,W$ fixed (and lying in some space which is not specified by the OP). Since $B$ appears in the second term only, we investigate it alone: $$ | Y - VB |_F ^2 = langle Y - VB , Y - VB rangle_F = |Y|_F ^2 + |V B |_F^2 - 2 langle Y , VB rangle_F. $$ For $B_1$ and $B_2$ matrices we get $$ begin{split} | Y - V (tB_1 + (1-t)B_2) |_F ^2 & =|Y|_F ^2 + |V (tB_1 + (1-t)B_2) |_F^2 - 2 langle Y , V(tB_1 + (1-t)B_2) rangle_F \ & le |Y|_F ^2 + t | VB_1 |_F ^2 + (1-t) |VB_2 |_F ^2 \ & - 2( t langle Y ,VB_1rangle_F + (1-t)langle Y ,VB_2rangle _F) end{split} $$where I have used the (bi)linearity of the inner product and the elementary fact that if $f,g $ are two convex functions with $g$ nondecreasing then $g circ f$ is a convex function; in our specific case $ g(x) = x^2$ and $f (X) = |X|_F$ (norms are convex).
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
$endgroup$
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I want to give an answer to the first question, expanding what was already hinted. The goal is to prove that the function $B mapsto f(V,W,B) $ is convex for every $V,W$ fixed (and lying in some space which is not specified by the OP). Since $B$ appears in the second term only, we investigate it alone: $$ | Y - VB |_F ^2 = langle Y - VB , Y - VB rangle_F = |Y|_F ^2 + |V B |_F^2 - 2 langle Y , VB rangle_F. $$ For $B_1$ and $B_2$ matrices we get $$ begin{split} | Y - V (tB_1 + (1-t)B_2) |_F ^2 & =|Y|_F ^2 + |V (tB_1 + (1-t)B_2) |_F^2 - 2 langle Y , V(tB_1 + (1-t)B_2) rangle_F \ & le |Y|_F ^2 + t | VB_1 |_F ^2 + (1-t) |VB_2 |_F ^2 \ & - 2( t langle Y ,VB_1rangle_F + (1-t)langle Y ,VB_2rangle _F) end{split} $$where I have used the (bi)linearity of the inner product and the elementary fact that if $f,g $ are two convex functions with $g$ nondecreasing then $g circ f$ is a convex function; in our specific case $ g(x) = x^2$ and $f (X) = |X|_F$ (norms are convex).
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
$endgroup$
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
add a comment |
$begingroup$
I want to give an answer to the first question, expanding what was already hinted. The goal is to prove that the function $B mapsto f(V,W,B) $ is convex for every $V,W$ fixed (and lying in some space which is not specified by the OP). Since $B$ appears in the second term only, we investigate it alone: $$ | Y - VB |_F ^2 = langle Y - VB , Y - VB rangle_F = |Y|_F ^2 + |V B |_F^2 - 2 langle Y , VB rangle_F. $$ For $B_1$ and $B_2$ matrices we get $$ begin{split} | Y - V (tB_1 + (1-t)B_2) |_F ^2 & =|Y|_F ^2 + |V (tB_1 + (1-t)B_2) |_F^2 - 2 langle Y , V(tB_1 + (1-t)B_2) rangle_F \ & le |Y|_F ^2 + t | VB_1 |_F ^2 + (1-t) |VB_2 |_F ^2 \ & - 2( t langle Y ,VB_1rangle_F + (1-t)langle Y ,VB_2rangle _F) end{split} $$where I have used the (bi)linearity of the inner product and the elementary fact that if $f,g $ are two convex functions with $g$ nondecreasing then $g circ f$ is a convex function; in our specific case $ g(x) = x^2$ and $f (X) = |X|_F$ (norms are convex).
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
$endgroup$
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
add a comment |
$begingroup$
I want to give an answer to the first question, expanding what was already hinted. The goal is to prove that the function $B mapsto f(V,W,B) $ is convex for every $V,W$ fixed (and lying in some space which is not specified by the OP). Since $B$ appears in the second term only, we investigate it alone: $$ | Y - VB |_F ^2 = langle Y - VB , Y - VB rangle_F = |Y|_F ^2 + |V B |_F^2 - 2 langle Y , VB rangle_F. $$ For $B_1$ and $B_2$ matrices we get $$ begin{split} | Y - V (tB_1 + (1-t)B_2) |_F ^2 & =|Y|_F ^2 + |V (tB_1 + (1-t)B_2) |_F^2 - 2 langle Y , V(tB_1 + (1-t)B_2) rangle_F \ & le |Y|_F ^2 + t | VB_1 |_F ^2 + (1-t) |VB_2 |_F ^2 \ & - 2( t langle Y ,VB_1rangle_F + (1-t)langle Y ,VB_2rangle _F) end{split} $$where I have used the (bi)linearity of the inner product and the elementary fact that if $f,g $ are two convex functions with $g$ nondecreasing then $g circ f$ is a convex function; in our specific case $ g(x) = x^2$ and $f (X) = |X|_F$ (norms are convex).
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
$endgroup$
I want to give an answer to the first question, expanding what was already hinted. The goal is to prove that the function $B mapsto f(V,W,B) $ is convex for every $V,W$ fixed (and lying in some space which is not specified by the OP). Since $B$ appears in the second term only, we investigate it alone: $$ | Y - VB |_F ^2 = langle Y - VB , Y - VB rangle_F = |Y|_F ^2 + |V B |_F^2 - 2 langle Y , VB rangle_F. $$ For $B_1$ and $B_2$ matrices we get $$ begin{split} | Y - V (tB_1 + (1-t)B_2) |_F ^2 & =|Y|_F ^2 + |V (tB_1 + (1-t)B_2) |_F^2 - 2 langle Y , V(tB_1 + (1-t)B_2) rangle_F \ & le |Y|_F ^2 + t | VB_1 |_F ^2 + (1-t) |VB_2 |_F ^2 \ & - 2( t langle Y ,VB_1rangle_F + (1-t)langle Y ,VB_2rangle _F) end{split} $$where I have used the (bi)linearity of the inner product and the elementary fact that if $f,g $ are two convex functions with $g$ nondecreasing then $g circ f$ is a convex function; in our specific case $ g(x) = x^2$ and $f (X) = |X|_F$ (norms are convex).
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
answered Nov 30 '18 at 16:52
gangrenegangrene
905514
905514
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
add a comment |
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
norm(V(tB1 + (1-t)B2, 'fro')^2 <= t*norm(VB1,'fro')^2 + (1-t)*norm(VB2,'fro')^2 ,why? This does not conform to the triangle inequality.
$endgroup$
– learn_truth
Dec 7 '18 at 2:42
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
$begingroup$
It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
$endgroup$
– gangrene
Dec 7 '18 at 13:10
add a comment |
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$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 7:17
$begingroup$
B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 9:29
$begingroup$
@Michal Adamaszek If the function is non-convex, we can use the anti-evidence method. But when only B is a matrix variable, the objective function is a convex function. Is it appropriate to use this anti-evidence method: (y-vb)^ 2?
$endgroup$
– learn_truth
Nov 28 '18 at 12:44
$begingroup$
@lilili My hint was not about proving general non-convexity. It was about how to prove convexity in B: Frobenius norm square is convex and the rest is linear.
$endgroup$
– Michal Adamaszek
Nov 28 '18 at 13:24
$begingroup$
@Michal Adamaszek The variable is a matrix, does the rule still apply?
$endgroup$
– learn_truth
Nov 29 '18 at 3:09