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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?

enter image description here

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  • $\begingroup$ 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? $\endgroup$ – StackTD May 23 at 11:31
  • $\begingroup$ It does stretch infinitely bounded by the dashed lines. Sorry for the confusion. $\endgroup$ – Parseval May 23 at 11:33
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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.

enter image description here

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

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.

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  • $\begingroup$ Wonderful explanation. Thanks! $\endgroup$ – Parseval May 23 at 12:36
  • $\begingroup$ You're welcome! $\endgroup$ – StackTD May 23 at 12:47
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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 $\lambda_1,\lambda_2\ge0$.

enter image description here

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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:

$\hspace{2cm}$enter image description here

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