Question-40

Find the matrix that projects every point in the plane onto the line \(\displaystyle x+2y=0\).

• \(\displaystyle \begin{bmatrix} 2 & -1\\ -1 & 2 \end{bmatrix}\)

• \(\displaystyle \begin{bmatrix} 4 & -2\\ -2 & 1 \end{bmatrix}\)

• \(\displaystyle \frac{1}{5}\begin{bmatrix} 2 & -1\\ -1 & 2 \end{bmatrix}\)

• \(\displaystyle \frac{1}{5}\begin{bmatrix} 4 & -2\\ -2 & 1 \end{bmatrix}\)

Answer

• \(\displaystyle \begin{bmatrix} 2 & -1\\ -1 & 2 \end{bmatrix}\)

• \(\displaystyle \begin{bmatrix} 4 & -2\\ -2 & 1 \end{bmatrix}\)

• \(\displaystyle \frac{1}{5}\begin{bmatrix} 2 & -1\\ -1 & 2 \end{bmatrix}\)

• \(\displaystyle \frac{1}{5}\begin{bmatrix} 4 & -2\\ -2 & 1 \end{bmatrix}\)

Solution

The unit vector along this line is \(\displaystyle v=\frac{1}{\sqrt{5}}( 2,-1)\). The projection of any point \(\displaystyle u\) along this line is therefore:

\[ \begin{equation*} \left( u^{T} v\right) v \end{equation*} \]

Expressing it slightly differently using the symmetry of dot-products and the associativity of matrix multiplication:

\[ \begin{equation*} \left( vv^{T}\right) u \end{equation*} \]

Therefore, the projection matrix is \(\displaystyle P=vv^{T}\) and is given by:

\[ \begin{equation*} P=\frac{1}{5}\begin{bmatrix} 4 & -2\\ -2 & 1 \end{bmatrix} \end{equation*} \]