Is the exponential of the trace convex?

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Take $a in mathbbR^+$ and $B,X in mathbbR^dtimes d$ where $B=B^T$ and $X = X^T$, $a$ and $B$ are fixed.



Is then the function
$$f(X) = ae^operatornametr(BX)$$ convex in X?



For $d=1$ this is definitely the case, but the introduction of the trace is what confounds me.




linear-algebra convex-analysis exponential-function trace




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    no need for any formulas; a convex function of an affine function is convex
    – LinAlg
    Aug 1 at 18:16














up vote
0
down vote

favorite












Take $a in mathbbR^+$ and $B,X in mathbbR^dtimes d$ where $B=B^T$ and $X = X^T$, $a$ and $B$ are fixed.



Is then the function
$$f(X) = ae^operatornametr(BX)$$ convex in X?



For $d=1$ this is definitely the case, but the introduction of the trace is what confounds me.




linear-algebra convex-analysis exponential-function trace




share|cite|improve this question

















  • 1




    no need for any formulas; a convex function of an affine function is convex
    – LinAlg
    Aug 1 at 18:16












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Take $a in mathbbR^+$ and $B,X in mathbbR^dtimes d$ where $B=B^T$ and $X = X^T$, $a$ and $B$ are fixed.



Is then the function
$$f(X) = ae^operatornametr(BX)$$ convex in X?



For $d=1$ this is definitely the case, but the introduction of the trace is what confounds me.




linear-algebra convex-analysis exponential-function trace




share|cite|improve this question













Take $a in mathbbR^+$ and $B,X in mathbbR^dtimes d$ where $B=B^T$ and $X = X^T$, $a$ and $B$ are fixed.



Is then the function
$$f(X) = ae^operatornametr(BX)$$ convex in X?



For $d=1$ this is definitely the case, but the introduction of the trace is what confounds me.





linear-algebra convex-analysis exponential-function trace





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 1 at 12:25









José Carlos Santos

112k1696172




112k1696172









asked Aug 1 at 11:57









NeedsToKnowMoreMaths

285




285







  • 1




    no need for any formulas; a convex function of an affine function is convex
    – LinAlg
    Aug 1 at 18:16












  • 1




    no need for any formulas; a convex function of an affine function is convex
    – LinAlg
    Aug 1 at 18:16







1




1




no need for any formulas; a convex function of an affine function is convex
– LinAlg
Aug 1 at 18:16




no need for any formulas; a convex function of an affine function is convex
– LinAlg
Aug 1 at 18:16










1 Answer
1






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up vote
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Yes, because if $X_1,X_2inmathbbR^dtimes d$ and $tin[0,1]$, thenbeginalignfbigl(tX_1+(1-t)X_2bigr)&=expbigl(operatornametrbigl(B.(tX_1+(1-t)X_2)bigr)bigr)\&=expbigl(operatornametrbigl(tBX_1+(1-t)BX_2bigr)bigr)\&=expbigl(toperatornametrbigl(BX_1bigr)+(1-t)operatornametrbigl(BX_2bigr)bigr)\&leqslant texpbigl(operatornametrbigl(BX_1bigr)bigr)+(1-t)expbigl(operatornametrbigl(BX_2bigr)bigr)\&=tf(X_1)+(1-t)f(X_2).endalign






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  • My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
    – user550103
    Aug 1 at 13:26











  • @user550103 I just used the fact that $exp$ is convex.
    – José Carlos Santos
    Aug 1 at 13:34










  • Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
    – NeedsToKnowMoreMaths
    Aug 2 at 13:01










  • @NeedsToKnowMoreMaths I'm glad I could help.
    – José Carlos Santos
    Aug 2 at 14:09










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1 Answer
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1 Answer
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active

oldest

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active

oldest

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oldest

votes








up vote
3
down vote



accepted










Yes, because if $X_1,X_2inmathbbR^dtimes d$ and $tin[0,1]$, thenbeginalignfbigl(tX_1+(1-t)X_2bigr)&=expbigl(operatornametrbigl(B.(tX_1+(1-t)X_2)bigr)bigr)\&=expbigl(operatornametrbigl(tBX_1+(1-t)BX_2bigr)bigr)\&=expbigl(toperatornametrbigl(BX_1bigr)+(1-t)operatornametrbigl(BX_2bigr)bigr)\&leqslant texpbigl(operatornametrbigl(BX_1bigr)bigr)+(1-t)expbigl(operatornametrbigl(BX_2bigr)bigr)\&=tf(X_1)+(1-t)f(X_2).endalign






share|cite|improve this answer





















  • My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
    – user550103
    Aug 1 at 13:26











  • @user550103 I just used the fact that $exp$ is convex.
    – José Carlos Santos
    Aug 1 at 13:34










  • Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
    – NeedsToKnowMoreMaths
    Aug 2 at 13:01










  • @NeedsToKnowMoreMaths I'm glad I could help.
    – José Carlos Santos
    Aug 2 at 14:09














up vote
3
down vote



accepted










Yes, because if $X_1,X_2inmathbbR^dtimes d$ and $tin[0,1]$, thenbeginalignfbigl(tX_1+(1-t)X_2bigr)&=expbigl(operatornametrbigl(B.(tX_1+(1-t)X_2)bigr)bigr)\&=expbigl(operatornametrbigl(tBX_1+(1-t)BX_2bigr)bigr)\&=expbigl(toperatornametrbigl(BX_1bigr)+(1-t)operatornametrbigl(BX_2bigr)bigr)\&leqslant texpbigl(operatornametrbigl(BX_1bigr)bigr)+(1-t)expbigl(operatornametrbigl(BX_2bigr)bigr)\&=tf(X_1)+(1-t)f(X_2).endalign






share|cite|improve this answer





















  • My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
    – user550103
    Aug 1 at 13:26











  • @user550103 I just used the fact that $exp$ is convex.
    – José Carlos Santos
    Aug 1 at 13:34










  • Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
    – NeedsToKnowMoreMaths
    Aug 2 at 13:01










  • @NeedsToKnowMoreMaths I'm glad I could help.
    – José Carlos Santos
    Aug 2 at 14:09












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Yes, because if $X_1,X_2inmathbbR^dtimes d$ and $tin[0,1]$, thenbeginalignfbigl(tX_1+(1-t)X_2bigr)&=expbigl(operatornametrbigl(B.(tX_1+(1-t)X_2)bigr)bigr)\&=expbigl(operatornametrbigl(tBX_1+(1-t)BX_2bigr)bigr)\&=expbigl(toperatornametrbigl(BX_1bigr)+(1-t)operatornametrbigl(BX_2bigr)bigr)\&leqslant texpbigl(operatornametrbigl(BX_1bigr)bigr)+(1-t)expbigl(operatornametrbigl(BX_2bigr)bigr)\&=tf(X_1)+(1-t)f(X_2).endalign






share|cite|improve this answer













Yes, because if $X_1,X_2inmathbbR^dtimes d$ and $tin[0,1]$, thenbeginalignfbigl(tX_1+(1-t)X_2bigr)&=expbigl(operatornametrbigl(B.(tX_1+(1-t)X_2)bigr)bigr)\&=expbigl(operatornametrbigl(tBX_1+(1-t)BX_2bigr)bigr)\&=expbigl(toperatornametrbigl(BX_1bigr)+(1-t)operatornametrbigl(BX_2bigr)bigr)\&leqslant texpbigl(operatornametrbigl(BX_1bigr)bigr)+(1-t)expbigl(operatornametrbigl(BX_2bigr)bigr)\&=tf(X_1)+(1-t)f(X_2).endalign







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 1 at 12:12









José Carlos Santos

112k1696172




112k1696172











  • My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
    – user550103
    Aug 1 at 13:26











  • @user550103 I just used the fact that $exp$ is convex.
    – José Carlos Santos
    Aug 1 at 13:34










  • Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
    – NeedsToKnowMoreMaths
    Aug 2 at 13:01










  • @NeedsToKnowMoreMaths I'm glad I could help.
    – José Carlos Santos
    Aug 2 at 14:09
















  • My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
    – user550103
    Aug 1 at 13:26











  • @user550103 I just used the fact that $exp$ is convex.
    – José Carlos Santos
    Aug 1 at 13:34










  • Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
    – NeedsToKnowMoreMaths
    Aug 2 at 13:01










  • @NeedsToKnowMoreMaths I'm glad I could help.
    – José Carlos Santos
    Aug 2 at 14:09















My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
– user550103
Aug 1 at 13:26





My apologies in advance... Just out of curiosity, the inequality you have invoked on the 4th row, is this some well-known? Jensen's inequality? If Jensen's inequality, then does it not assume that the function is convex/concave (so kind of chicken or egg-like problem)? or one could also show that the $f^primeprime(X) > 0$ for the convexity, right?
– user550103
Aug 1 at 13:26













@user550103 I just used the fact that $exp$ is convex.
– José Carlos Santos
Aug 1 at 13:34




@user550103 I just used the fact that $exp$ is convex.
– José Carlos Santos
Aug 1 at 13:34












Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
– NeedsToKnowMoreMaths
Aug 2 at 13:01




Thanks! I had sort of convinced myself of this because the trace is affine, but it's good to see it in a few lines.
– NeedsToKnowMoreMaths
Aug 2 at 13:01












@NeedsToKnowMoreMaths I'm glad I could help.
– José Carlos Santos
Aug 2 at 14:09




@NeedsToKnowMoreMaths I'm glad I could help.
– José Carlos Santos
Aug 2 at 14:09












 

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