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I am reading Convex Optimization written by Stephen Boyd. In page 27 of chapter 2, there is an example said 'Any subspace is affine, and a convex cone(hence convex).' Can anybody explain to me why this is true?

BioCoder
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    Have you tried applying the definition of what a convex cone is to see if a subspace satisfies it? That usually does the trick! – Mariano Suárez-Álvarez Feb 28 '15 at 10:07
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    Yes I did. Actually, what confused me is that if any subspace is a convex cone, isn't any space a convex cone ? Because any space is a subspace of another space whose dimension is higher, isn't it? – BioCoder Mar 01 '15 at 05:24

2 Answers2

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First, what basically distinguishes the definitions of convex, affine and cone, is the domain of the coefficients and the constraints that relate them.

Let us starts by the first part: any subspace is affine, which means, if we have:

$x_1, x_2 \in V$, where $V$ is a subspace; therefore any linear combination of these two vectors must lie in $V$. That is, if we have two coefficients $\theta_1, \theta_2 \in \mathcal{R}$, then, $\theta_1x_1 + \theta_2x_2 \in V$.

The definition of affine sets tells us if $x_1,x_2$ are in an affine set, their linear combination must also lie in the same set, with the condition the coefficients must sum to 1, that is $\theta_1 + \theta_2 = 1$. Now, assume we have chosen $\theta_2 = 1- \theta_1$, therefore the combination $\theta_1x_1 + (1 - \theta_1)x_2 \in V$

Therefore any subspace is affine, since we have the freedom to choose the coefficients to sum to 1.

Now why a subspace is a convex cone.

Notice that, if we choose the coeficientes $\theta_1, \theta_2 \in \mathcal{R}_+$, we actually define a cone, and if the coefficients sum to 1, it is convex, therefore it is a convex cone.

Faroq AL-Tam
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because a linear subspace contains all multiples of its elements as well as all linear combinations (in particular convex ones). Hence it is a convex cone.

daw
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