What is the tangent cone of the nonnegative orthant?

Question

Definition:

$T_c(x)$ is the tangent cone of $C$ at point $x \in C$

$T_c(x) = closure\{z \in \mathbb{R}^n, z = k(y - x), \forall \thinspace y \in C, k \geq 0 \}$

And $T_c(x)$ is a closed convex cone

Consider the nonnegative orthant denoted as $C$, the nonegative orthant is obviously a closed convex cone.

Consider $x = [1,1]$

Then take the boundary vectors, subtracting, yields $[1,0] - [1,1] = [0,-1]$, and $[0,1]-[1,1] = [-1,0]$

Then take any other vectors such as $[0.5, 1]$, or $[1, 0.6]$, subtracting off $[1,1]$, I get

A set which is obviously not convex

What am I doing wrong here? How should I construct the tangent cone of the nonnegative orthant and what does it look like?

Answer 1

It is known that (see here, for example) if $X$ is a finite-dimensional vector space and $C$ is a convex subset of $X$, then $T_C(x) = X$ for any $x \in C^\circ$. So in your particular case $T_C(x) = X = \mathbb{R}^2$.

Answer 2

You have to compute the difference $y - x$ for all $y \in C$.

In particular, since $x$ belongs to the interior of $C$, you get $T_C(x) = \mathbb{R}^2$.