# Draw region given set of points.

Can someone explain how

$$\{\mathbf{x}\in\mathbb{R}^2 \ | \ \mathbf{x}=\begin{pmatrix}2\\2\end{pmatrix} + \lambda_1\begin{pmatrix}-1\\3\end{pmatrix}+\lambda_2\begin{pmatrix}2\\1\end{pmatrix};\lambda_1,\lambda_2>0\}$$

is the green area below?

• Either the lambda's need to be bounded above (to get the green triangle as plotted), or only a part of the green area is plotted and you should interpret it as stretching out infinitely further to the top right. Could you clarify? – StackTD May 23 at 11:31
• It does stretch infinitely bounded by the dashed lines. Sorry for the confusion. – Parseval May 23 at 11:33

A picture says more than a bunch of words but since making a digital one would take a lot more time, here's an old school drawing!

Start from $$(2,2)$$ in blue:

• adding $$(-1,3)$$ takes you to the green dot, adding any (positive!) scalar multiple of $$(-1,3)$$ lets you move on the green (half-)line: this is $$(2,2)+\lambda_1(-1,3)$$ with $$\lambda_1 >0$$;

• adding $$(2,1)$$ takes you to the red dot, adding any (positive!) scalar multiple of $$(2,1)$$ lets you move on the red (half-)line: this is $$(2,2)+\lambda_2(2,1)$$ with $$\lambda_2 >0$$;

• adding a positive scalar multiple of $$(-1,3)$$ and $$(2,1)$$ corresponds to the vector addition of the previous two cases (with $$(2,2)$$ as the center), indicated by the dashed black lines and the final black dot for one specific choice of $$\lambda_1$$ and $$\lambda_2$$.

Now that black dot is where you arrive for a specific choice of $$\lambda_1$$ and $$\lambda_2$$, letting both scalars take all positive values will let the black dot move through the entire marked region.

$$\{\mathbf{x}\in\mathbb{R}^2 \ | \ \mathbf{x}=\begin{pmatrix}2\\2\end{pmatrix} + \lambda_1\begin{pmatrix}-1\\3\end{pmatrix}+\lambda_2\begin{pmatrix}2\\1\end{pmatrix};\lambda_1,\lambda_2>0\} \tag{\star}$$

This parametrization of the shaded region works in two ways:

• to any point $$\mathbf{x}$$ in the shaded region, correspond (unique) values of $$\lambda_1$$ and $$\lambda_2$$ such that $$\mathbf{x}$$ can be written in the form as given in $$(\star)$$;

• taking any two positive values for $$\lambda_1$$ and $$\lambda_2$$ in $$(\star)$$ will result in a point $$\mathbf{x}$$ which is located in the shaded region.

• Wonderful explanation. Thanks! – Parseval May 23 at 12:36
• You're welcome! – StackTD May 23 at 12:47

Set $$\lambda_1=\lambda_2=0$$ and you get the corner point $$(2,2)$$.

Now increase $$\lambda_1$$ and you will follow a linear edge, through $$(2-1,2+3)$$ (among others).

Restart from the corner and increase $$\lambda_2$$: another line, through $$(2+2,2+1)$$.

These two half-lines delimit an infinite sector of the plane such that $$\lambda_1,\lambda_2\ge0$$.

Alternatively: $$x=\begin{pmatrix}x_1\\ x_2\end{pmatrix}=\begin{pmatrix}2-\lambda_1+2\lambda_2\\ 2+3\lambda_1+\lambda_2\end{pmatrix} \Rightarrow \begin{cases}\lambda_1=2-x_1+2\lambda_2\\ x_2=2+3(2-x_1+2\lambda_2)+\lambda_2=-3x_1+8+7\lambda_2\end{cases}$$ Similarly: $$x=\begin{pmatrix}x_1\\ x_2\end{pmatrix}=\begin{pmatrix}2-\lambda_1+2\lambda_2\\ 2+3\lambda_1+\lambda_2\end{pmatrix} \Rightarrow \begin{cases}x_1=2-\lambda_1+2(x_2-2-3\lambda_1)\Rightarrow x_2=\frac12x_1+1+\frac72\lambda_1\\ \lambda_2=x_2-2-3\lambda_1\end{cases}$$ Hence, for $$\lambda_1>0,\lambda_2>0$$, it is: $$\begin{cases}x_2=-3x_1+8+7\lambda_2>-3x_1+8\\ x_2=\frac12x_1+1+\frac72\lambda_1>\frac12x_1+1\end{cases}$$ The feasible region is:

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