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- Apr 14, 2013

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To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?

- Thread starter mathmari
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- Thread starter
- #1

- Apr 14, 2013

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To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?

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- #2

- Mar 5, 2012

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Hi!

To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?

Is the Hessian matrix positive semi-definite?

Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?

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- Apr 14, 2013

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for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $ H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}] $. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?Hi!

Is the Hessian matrix positive semi-definite?

Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?

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- #4

- Mar 5, 2012

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Yep.for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $ H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}] $. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite.

(Although you should leave out the absolute value symbols for $D1$. )

Positive definite requires $>0$, which is not the case.But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?

Similarly negative definite requires $<0$, which is also not the case.

So if the hessian matrix is the zero matrix it is neither positive definite nor negative definite.

However, it

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- #5

- Apr 14, 2013

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so do we conlude that the function is both concave and convex??However, itisboth positive semi-definite and negative semi-definite.

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Yes.so do we conlude that the function is both concave and convex??

Note that it is neither

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- #7

- Apr 14, 2013

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Ok! Thank you!Yes.

Note that it is neitherstrictlyconvex, norstrictlyconcave.

- Feb 15, 2012

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- #9

- Apr 14, 2013

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Do you mean that this is the technical term that a function is both concave and convex?