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Show that a translated subspace, i.e., $D=\{Ev+d \mid v \in \mathbb{R}^p\} \subseteq \mathbb{R}^n$ is a convex set where $E \in \mathbb{R}^{n \times p}$ and $d \in \mathbb{R}^n$.

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    What have you tried? Have you written down the definition of a convex set and tried to check if $D$ satisfies the conditions? – angryavian Dec 03 '18 at 04:37
  • Yes. I have considered $x_1$ and $x_2$ in $D$. $x_1=Ev_1+d$ and $x_2=Ev_2+d$. Multiply by $\lambda$ and sum. I am not sure if I can write $x_1=Ev_1+d$ and $x_2=Ev_2+d$ or not? –  Dec 03 '18 at 04:55
  • Yes, your approach is good. – angryavian Dec 03 '18 at 04:56
  • @Saeed please add the information from the comment to the question. Try and explain what you are doing and where you are stuck. – DRF Dec 03 '18 at 09:44

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If $x_1,x_2 \in D$, then $x_1 = Ev_1+d, x_2 = Ev_2+d$ for some $v_1,v_2 \in \mathbb{R}^p$, respectively. Then, $D$ is convex if $$x_3 = \lambda x_1 + (1-\lambda)x_2 \in D \text{ for all }\lambda\in(0,1)$$ i.e. there exists a $v_3\in\mathbb{R}^p$ such that $x_3 = Ev_3+d$. Substituting in the expressions for $x_1,x_2$, we see that

$$ \lambda(Ev_1+d) + (1-\lambda)(Ev_2+d) = E\lambda v_1 + \lambda d + E(1-\lambda)v_2 + (1-\lambda)d = E(\lambda v_1 + (1-\lambda)v_2) + d $$

Since $\lambda v_1 + (1-\lambda)v_2 \in \mathbb{R}^p$ for any $v_1,v_2\in\mathbb{R}^p$ and $\lambda\in(0,1)\subset\mathbb{R}$, we let $v_3 = \lambda v_1 + (1-\lambda)v_2$, and so it follows that $D$ is convex.

Alex Jones
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