Question

Let \({\bf{u}}\) and \({\bf{v}}\) be vectors in\({\mathbb{R}^{\bf{n}}}\). It can be shown that the set \({\bf{P}}\) of all points in the parallelogram determined by \({\bf{u}}\) and \({\bf{v}}\) has the form \({\bf{av}} + {\bf{bv}}\), for \({\bf{0}} \le {\bf{a}} \le {\bf{1}}\), \({\bf{0}} \le {\bf{b}} \le {\bf{1}}\). Let \({\bf{T}}:{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{n}}}\) be a linear transformation. Explain why the image of a point in \({\bf{P}}\) under the transformation \({\bf{T}}\) lies in the parallelogram determined by \({\bf{T}}\left( {\bf{u}} \right)\) and \({\bf{T}}\left( {\bf{v}} \right)\).

Short answer

The image of any point in \(P\) is of the form \(aT\left( u \right) + bT\left( v \right)\) for\(a,b \in \left( {0,1} \right)\). Hence, the image of any point in \(P\) under the transformation \(T\) lies in the parallelogram determined by \(T\left( u \right)\) and \(T\left( v \right)\).

Step-by-step solution

Use the definition of \({\bf{P}}\)

The given parallelogram \(P\) can be written as \(P = \left\{ {au + bv:0 \le a \le 1,0 \le a \le 1} \right\}\).

Determine the image of a point in \({\bf{P}}\) under the transformation \({\bf{T}}\)

Let \(x \in P\). Then

\(\begin{aligned} T\left( x \right) &= T\left( {au + bv} \right)\\ &= T\left( {au} \right) + T\left( {bv} \right)\\T\left( x \right) &= aT\left( u \right) + bT\left( v \right)\,\,\,\,\,\,\,{\rm{ }}a,b \in \left( {0,1} \right)\end{aligned}\)

Conclusion

So, the image of any point in \(P\) is of the form \(aT\left( u \right) + bT\left( v \right)\) for \(a,b \in \left( {0,1} \right)\). This implies that the image of any point in \(P\) under the transformation \(T\) lies in the parallelogram determined by \(T\left( u \right)\) and \(T\left( v \right)\).