Let \(L \subset \mathbb{R}^2\) be the line spanned by the vector \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\). Let \(P \subset \mathbb{R}^2\) be the line defined by the equation \(x + 2y = 0\). Find the standard matrix for the linear transformation \(T : \mathbb{R}^2 \to \mathbb{R}^2\) such that \(T(v) = 0\) for \(v \in L\) and \(T(v) = v\) for \(v \in P\).
<1>4. Therefore the standard matrix of \(T\) is \[ \boxed{ [T]=\frac15 \begin{pmatrix} 4&-2\\ -2&1 \end{pmatrix}.} \]
Solution
1. The vectors\[ \ell=\begin{pmatrix}1\\2\end{pmatrix}, \qquad p=\begin{pmatrix}2\\-1\end{pmatrix} \]form a basis of \(\mathbb R^2\), with \(\ell\in L\) and \(p\in P\).
The vector \(p\) satisfies \(2+2(-1)=0\), so \(p\in P\). Also \[ \det\begin{pmatrix}1&2\\2&-1\end{pmatrix}=-5\ne0, \] so \(\ell,p\) are linearly independent.
2. Every vector \((x,y)^t\) has a unique decomposition\[ \begin{pmatrix}x\\y\end{pmatrix} =a\ell+bp, \qquad b=\frac{2x-y}{5}. \]
The equations are \[ x=a+2b,\qquad y=2a-b. \] Substituting \(a=x-2b\) into the second equation gives \[ y=2x-5b, \] so \(b=(2x-y)/5\).
3. Since \(T\) vanishes on \(L\) and is the identity on \(P\),\[ T\begin{pmatrix}x\\y\end{pmatrix} =b p =\frac{2x-y}{5}\begin{pmatrix}2\\-1\end{pmatrix}. \]
Expanding 3 gives \[ T(x,y)=\left(\frac{4x-2y}{5},\frac{-2x+y}{5}\right), \] whose coefficient matrix is the displayed matrix.