Let \({\bf{u}}\) and \({\bf{v}}\) be linearly independent vectors in \({\mathbb{R}^{\bf{3}}}\), and let \({\bf{P}}\) be the plane through \({\bf{u}}\),\({\bf{v}}\),and \({\bf{0}}\). The parametric equation of \({\bf{P}}\) is \({\bf{x}} = {\bf{su}} + {\bf{tv}}\) (with\({\bf{s}}\),\({\bf{t}}\) in \(\mathbb{R}\)). Show that a linear transformation \({\bf{T}}:{\mathbb{R}^{\bf{3}}} \to {\mathbb{R}^{\bf{3}}}\) maps \({\bf{P}}\) onto a plane through\({\bf{0}}\), or onto a line through\({\bf{0}}\), or onto just the origin in \({\mathbb{R}^{\bf{3}}}\). What must be true about \({\bf{T}}\left( {\bf{u}} \right)\) and \({\bf{T}}\left( {\bf{v}} \right)\) in order for the image of the plane \({\bf{P}}\) to be a plane?

\(\begin{aligned} T\left( P \right) &= T\left( {su + tv} \right)\\ &= T\left( {su} \right) + T\left( {tv} \right)\\T\left( p \right) &= sT\left( u \right) + tT\left( v \right)\,\,\,\,\,\,\,\,\,\,\,s,t \in \mathbb{R}\end{aligned}\)

This implies \(T\left( P \right)\) is a set of all linear combinations of \(T\left( u \right)\) and \(T\left( v \right)\).

Whenever \(T\left( u \right)\) and \(T\left( v \right)\) are linearly independent, they form a basis for \(T\left( P \right)\).

This implies \(T\left( P \right)\) is a plane through 0, \(T\left( u \right)\), and \(T\left( v \right)\).

Suppose \(T\left( u \right)\) and \(T\left( v \right)\) are linearly dependent with \(T\left( u \right) = T\left( v \right) = 0\).

Then, \(T\left( P \right) = s \cdot 0 + t \cdot 0 = 0\). This implies \(T\left( P \right)\) is just the origin in \({\mathbb{R}^3}\).

Suppose both \(T\left( u \right)\) and \(T\left( v \right)\) are not zeros. Then, either \(T\left( P \right) = sT\left( u \right)\) or \(T\left( P \right) = t{\rm{T}}\left( v \right)\).

This implies \(T\left( P \right)\) is a straight line through \(0\).

We know that \(T\left( P \right) = {\rm{Span}}\left\{ {T\left( u \right),T\left( v \right)} \right\}\).

Suppose \(au + bv = 0\) then \(a = b = 0\). Since \(u\) and \(v\) are linearly independent,

\(\begin{array}{c}T\left( {au + bv} \right) = T\left( 0 \right)\\T\left( {au} \right) + T\left( {bv} \right) = 0\\aT\left( u \right) + bT\left( v \right) = 0\end{array}\)

Here, \(a = b = 0\). This implies \(T\left( u \right)\) and \(T\left( v \right)\) are linearly independent.

Thus, \(T\left( u \right)\) and \(T\left( v \right)\) should be a basis for \(T\left( P \right)\). Hence, \(T\left( P \right)\) must be a plane.