Question

Let y = (2 6) and u = (7 1). Write y as the sum of a vector in Span(u) and a vector orthogonal to .

Answer 1

The question is missing. Here is the complete question.

Let y = $\left[\begin{array}{ccc}2\\6\end{array}\right]$ and u = $\left[\begin{array}{ccc}7\\1\end{array}\right]$. Write y as the sum of a vector in Span(u) and a vector orthogonal to u.

Answer: y = $\left[\begin{array}{ccc}\frac{21}{10} \\ \frac{3}{10} \end{array}\right] + \left[\begin{array}{ccc}\frac{-1}{10}\\ \frac{57}{10} \end{array}\right]$

Step-by-step explanation: The sum of vectors is given by

y =  $y_{1}$ + z

where  $y_{1}$ is in Span(u);

vector z is orthogonal to it;

First you have to compute the orthogonal projection $y_{1}$ of y:

$y_{1}$ = proj y = $\frac{y.u}{u.u}.u$

Calculating orthogonal projection:

$\left[\begin{array}{c}2\\6\end{array}\right]$.$\left[\begin{array}{c}7\\1\end{array}\right]$ = $\left[\begin{array}{c}9\\6\end{array}\right]$

$\left[\begin{array}{c}7\\1\end{array}\right]$.$\left[\begin{array}{c}7\\1\end{array}\right]$ = $\left[\begin{array}{c}49\\1\end{array}\right]$

$y_{1} = \frac{9+6}{49+1}.u$

$y_{1} = \frac{15}{50}.u$

$y_{1} = \frac{3}{10}.u$

$y_{1} = \frac{3}{10}.\left[\begin{array}{c}7\\1\end{array}\right]$

$y_{1} = \left[\begin{array}{c}\frac{21}{10} \\\frac{3}{10} \end{array}\right]$

Calculating vector z:

z = y - $y_{1}$

z = $\left[\begin{array}{c}2\\6\end{array}\right] - \left[\begin{array}{c}\frac{21}{10} \\\frac{3}{10} \end{array}\right]$

z = $\left[\begin{array}{c}\frac{-1}{10} \\\frac{57}{10} \end{array}\right]$

Writing y as the sum:

$y = \left[\begin{array}{c}\frac{21}{10} \\\frac{3}{10} \end{array}\right] + \left[\begin{array}{c}\frac{-1}{10} \\\frac{57}{10} \end{array}\right]$