Question

Find the matrix A of the orthogonal projection onto the line L in R2 that consists of all scalar multiples of the vector [12].

Answer 1

Answer:

$A=\frac{1}{5}\left[\begin{array}{ccc}1&2\\2&4\end{array}\right]$

Step-by-step explanation:

Given vector

u1=$\left[\begin{array}{ccc}1\\2\end{array}\right]$

u2=$\left[\begin{array}{ccc}2\\-1\end{array}\right]$

Using orthonormal basis of R2

y={v1,v2}

$||u_1||=||u_2||=\sqrt{1^2+2^2}=\sqrt{5}$

$y=${$\frac{1}{\sqrt{5}}$$\left[\begin{array}{ccc}1\\2\end{array}\right]$,$\frac{1}{\sqrt{5}}\left[\begin{array}{ccc}2\\-1\end{array}\right]$}

We know that T(v1)=u,T(V2)=0

Let

Q(y)=$\frac{1}{\sqrt{5}}$$\left[\begin{array}{ccc}1&2\\2&-1\end{array}\right]$

and

$T(y)=\left[\begin{array}{ccc}1&0\\0&0\end{array}\right]$

$A=Q(y)\cdot T(y)\cdot Q(y)$

$A=\frac{1}{\sqrt{5}}\cdot\left[\begin{array}{ccc}1&2\\2&-1\end{array}\right]\cdot \left[\begin{array}{ccc}1&0\\0&0\end{array}\right]\cdot \frac{1}{\sqrt{5}} \left[\begin{array}{ccc}1&2\\2&-1\end{array}\right]$

$A=\frac{1}{5}\left[\begin{array}{ccc}1&0\\2&0\end{array}\right] \cdot \left[\begin{array}{ccc}1&2\\2&-1\end{array}\right]$

$A=\frac{1}{5}\left[\begin{array}{ccc}1&2\\2&4\end{array}\right]$