# Prove that By is a convex set

• Oct 18th 2011, 09:36 AM
Lolyta
Prove that By is a convex set
Hello.
Definition: A is convex if and only if $\displaystyle \forall{x_1,x_2\in{A}}$ the segment which unite them $\displaystyle (\lambda x_1 + (1-\lambda)x_2 ) \in{A}; 0\leq{\lambda\leq{1}}$.
So, let consider the set of vectors $\displaystyle By$ where B is a n x m matrix and $\displaystyle y = (y_1,y_2,...,y_m): y_i\geq{0}$ is an m-vector : $\displaystyle y_1+y_2+...+y_m=1$, prove that By is a convex set.

Thanks a lot.
• Oct 18th 2011, 10:10 AM
girdav
Re: Prove that By is a convex set
What did you try?
• Oct 18th 2011, 10:45 AM
Lolyta
Re: Prove that By is a convex set
Quote:

Originally Posted by girdav
What did you try?

Ok,the main problem is that considering $\displaystyle x_1,x_2\in{By}$, I don't know how $\displaystyle x_1,x_2$ are. I don't know if you understand me.(Worried)
• Oct 18th 2011, 10:51 AM
girdav
Re: Prove that By is a convex set
Let $\displaystyle v_1$ and $\displaystyle v_2$ in this set. We can write $\displaystyle v_1=Bx$ and $\displaystyle v_2=By$, where $\displaystyle x=(x_1,\ldots,x_m)$ and $\displaystyle y=(y_1,\ldots,y_n)$ are such that $\displaystyle x_j,y_j\geq 0$ for all $\displaystyle j$ and $\displaystyle \sum_{j=1}^mx_j=\sum_{j=1}^mx_j=1$. Take $\displaystyle \alpha\in\left[0,1\right]$. Then $\displaystyle \alpha v_1+(1-\alpha)v_2=\alpha Bx+(1-\alpha)By=B(\alpha x+(1-\alpha)y)$. What about $\displaystyle \alpha x+(1-\alpha)y$?
• Oct 18th 2011, 11:13 AM
Lolyta
Re: Prove that By is a convex set
wow! http://latex.codecogs.com/png.latex?...1-%5Calpha%29y =$\displaystyle \alpha\sum_{j=1}^mx_j + (1-\alpha)\sum_{j=1}^yx_j= \alpha + 1 - \alpha = 1$ which implies that http://latex.codecogs.com/png.latex?...1-%5Calpha%29y is in By.

Thank you!!!!(Itwasntme)