Prove that By is a convex set

**Lolyta** · October 18th 2011, 09:36 AM

Hello.
Definition: A is convex if and only if $\forall{x_1,x_2\in{A}}$ the segment which unite them $(\lambda x_1 + (1-\lambda)x_2 ) \in{A}; 0\leq{\lambda\leq{1}}$ .
So, let consider the set of vectors $By$ where B is a n x m matrix and $y = (y_1,y_2,...,y_m): y_i\geq{0}$ is an m-vector : $y_1+y_2+...+y_m=1$ , prove that By is a convex set.

Thanks a lot.

**girdav** · October 18th 2011, 10:10 AM

What did you try?

**Lolyta** · October 18th 2011, 10:45 AM

Originally Posted by girdav

What did you try?

Ok,the main problem is that considering $x_1,x_2\in{By}$ , I don't know how $x_1,x_2$ are. I don't know if you understand me.

**girdav** · October 18th 2011, 10:51 AM

Let $v_1$ and $v_2$ in this set. We can write $v_1=Bx$ and $v_2=By$ , where $x=(x_1,\ldots,x_m)$ and $y=(y_1,\ldots,y_n)$ are such that $x_j,y_j\geq 0$ for all $j$ and $\sum_{j=1}^mx_j=\sum_{j=1}^mx_j=1$ . Take $\alpha\in\left[0,1\right]$ . Then $\alpha v_1+(1-\alpha)v_2=\alpha Bx+(1-\alpha)By=B(\alpha x+(1-\alpha)y)$ . What about $\alpha x+(1-\alpha)y$ ?

**Lolyta** · October 18th 2011, 11:13 AM

wow! $\alpha x+(1-\alpha)y$ = $\alpha\sum_{j=1}^mx_j + (1-\alpha)\sum_{j=1}^yx_j= \alpha + 1 - \alpha = 1$ which implies that $\alpha x+(1-\alpha)y$ is in By.

Thank you!!!!

Thread: Prove that By is a convex set

LinkBack

Thread Tools

Display

Prove that By is a convex set

Re: Prove that By is a convex set

Re: Prove that By is a convex set

Re: Prove that By is a convex set

Re: Prove that By is a convex set

Similar Math Help Forum Discussions

Prove a convex function is continuous

Inequality to prove involving convex notions

Prove A Function Is Convex?

Prove a set is convex algebraically

If a function f is convex on (a,b) , prove that f is continuous on (a,b)

Search Tags