# A linear map from $R^3$ into $R^2$

Suppose $a\in (0,1)$ and

$$X=\{(x_1,x_2,x_3)\in R^3: a x_1+(1-a) x_2+ x_3\leq 3, x_i\geq 1, i=1,2,3.\}.$$

Define a linear map $\Gamma$ by $(x_1,x_2,x_3)\to (a x_1+(1-a) x_2, x_3)$ .

Do we have $\Gamma(X)$ equal to

$$Y=\{(y_1,y_2)\in R^2: y_1+ y_2\leq 3, y_1\geq 1, y_2\geq 1\}$$.

or $\Gamma(X)\subsetneq Y$?

• Any thoughts on how to attack such problems generally? Linear transformations are also convex functions, so this tells us that we can determine the image of a convex polyhedron in $\mathbb{R}^3$ by taking the convex closure of the images of the extremal points (vertices). – hardmath Jun 26 '16 at 13:55

For $(x_0, x_3)$ in $Y$, the point $(x_0, x_0, x_3)$ is in X and maps to $(x_0, x_3)$.
$(1)\ \Gamma(x_1,x_2,x_3)=(a x_1+(1-a) x_2, x_3)=(y_1,y_2)\rightarrow y_1=a x_1+(1-a) x_2\ \text{and} \ y_2=x_3.$
$(2)\ a x_1+(1-a) x_2+ x_3\leq 3 \ \rightarrow y_1+y_2\leq 3.$
$(3)\ x_i\geq 1,\ i=1,2,3 \ \rightarrow y_1=a x_1+(1-a) x_2\geq a+(1-a)=1 \ \ \text{and}\ \ y_2=x_3\geq 1 .$
So, according to $(1),\ (2),$ and $(3)$, it is concluded that $$\Gamma(X)=Y$$