How to prove a matrix function is convex or nonconvex?












0














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?










share|cite|improve this question
























  • 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.
    – José Carlos Santos
    Nov 28 at 7:17










  • B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
    – Michal Adamaszek
    Nov 28 at 9:29










  • @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?
    – learn_truth
    Nov 28 at 12:44










  • @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.
    – Michal Adamaszek
    Nov 28 at 13:24










  • @Michal Adamaszek The variable is a matrix, does the rule still apply?
    – learn_truth
    Nov 29 at 3:09
















0














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?










share|cite|improve this question
























  • 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.
    – José Carlos Santos
    Nov 28 at 7:17










  • B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
    – Michal Adamaszek
    Nov 28 at 9:29










  • @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?
    – learn_truth
    Nov 28 at 12:44










  • @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.
    – Michal Adamaszek
    Nov 28 at 13:24










  • @Michal Adamaszek The variable is a matrix, does the rule still apply?
    – learn_truth
    Nov 29 at 3:09














0












0








0







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?










share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




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edited Dec 9 at 12:57

























asked Nov 28 at 7:11









learn_truth

12




12












  • 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.
    – José Carlos Santos
    Nov 28 at 7:17










  • B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
    – Michal Adamaszek
    Nov 28 at 9:29










  • @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?
    – learn_truth
    Nov 28 at 12:44










  • @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.
    – Michal Adamaszek
    Nov 28 at 13:24










  • @Michal Adamaszek The variable is a matrix, does the rule still apply?
    – learn_truth
    Nov 29 at 3:09


















  • 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.
    – José Carlos Santos
    Nov 28 at 7:17










  • B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
    – Michal Adamaszek
    Nov 28 at 9:29










  • @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?
    – learn_truth
    Nov 28 at 12:44










  • @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.
    – Michal Adamaszek
    Nov 28 at 13:24










  • @Michal Adamaszek The variable is a matrix, does the rule still apply?
    – learn_truth
    Nov 29 at 3:09
















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.
– José Carlos Santos
Nov 28 at 7:17




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.
– José Carlos Santos
Nov 28 at 7:17












B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
– Michal Adamaszek
Nov 28 at 9:29




B appears in only one place. So all you need to know is that the square of Frobenius norm is convex.
– Michal Adamaszek
Nov 28 at 9:29












@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?
– learn_truth
Nov 28 at 12:44




@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?
– learn_truth
Nov 28 at 12:44












@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.
– Michal Adamaszek
Nov 28 at 13:24




@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.
– Michal Adamaszek
Nov 28 at 13:24












@Michal Adamaszek The variable is a matrix, does the rule still apply?
– learn_truth
Nov 29 at 3:09




@Michal Adamaszek The variable is a matrix, does the rule still apply?
– learn_truth
Nov 29 at 3:09










1 Answer
1






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1














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).






share|cite|improve this answer





















  • 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.
    – learn_truth
    Dec 7 at 2:42










  • It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
    – gangrene
    Dec 7 at 13:10











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1














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).






share|cite|improve this answer





















  • 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.
    – learn_truth
    Dec 7 at 2:42










  • It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
    – gangrene
    Dec 7 at 13:10
















1














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).






share|cite|improve this answer





















  • 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.
    – learn_truth
    Dec 7 at 2:42










  • It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
    – gangrene
    Dec 7 at 13:10














1












1








1






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).






share|cite|improve this answer












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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 16:52









gangrene

915514




915514












  • 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.
    – learn_truth
    Dec 7 at 2:42










  • It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
    – gangrene
    Dec 7 at 13:10


















  • 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.
    – learn_truth
    Dec 7 at 2:42










  • It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
    – gangrene
    Dec 7 at 13:10
















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.
– learn_truth
Dec 7 at 2:42




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.
– learn_truth
Dec 7 at 2:42












It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
– gangrene
Dec 7 at 13:10




It is justified below, norms are convex and composition between nondecreasing convex and convex gives you another convex function.
– gangrene
Dec 7 at 13:10


















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