Question

Verify that (41.uz) is an orthogonal set, and then find the orthogonal projection of y onto Span{41.42}- y = 1 0 To verify that {0, 42} is an orthogonal set, find u, '42. u U2 - 0 (Simplify your answer) The projection of y onto Span{u, uz} is (Simplify your answers)

Answer 1

We know that if  
{u, uz
is an orthogonal set, then the dot product of $u_{1}$ and $u_{2}$ is 0 i.e. $u_{1}\cdot u_{2}=0$ .

In the given question,

4 u1 5 and u2 0 -5 4 0

$So,\:u_{1}\cdot u_{2}=(4\times -5)+(5\times 4)+(0\times 0)=-20+20+0=0\\ \Rightarrow u_{1}\cdot u_{2}=0\\ \Rightarrow u_{1} \: and\: u_{2} \: are\:orthogonal.$$\Rightarrow \left \{ u_{1},u_{2} \right \} \:is\:an\:orthogonal \:set.$

Now, we will find the orthogonal projection of y onto $Span\left \{ \textbf{u}_{1},\textbf{u}_{2} \right \}$ .

Let the orthogonal projection of y onto  $Span\left \{ \textbf{u}_{1},\textbf{u}_{2} \right \}$ is  $\textbf{u}$.

$Since,\: u\: \epsilon \:Span\left \{ u_{1},u_{2} \right \}, \\ Let \:\textbf{u} = a_{1}\textbf{u}_{1}+a_{2}\textbf{u}_{2} \:for\: some \:scalar\:a_{1},a_{2}.$

Now, we have to find the values of $a_{1}$ and $a_{2}$ .

$Using \:the \:formula\:\:\: a_{1}= \frac{\textbf{y}\cdot \textbf{u}_{1}}{\textbf{u}_{1}\cdot \textbf{u}_{1}},\:we\:get\:,$

$a_{1}= \frac{\textbf{y}\cdot \textbf{u}_{1}}{\textbf{u}_{1}\cdot \textbf{u}_{1}}\\ \Rightarrow a_{1}=\frac{(5\times4)+(3\times5)+(-1\times0)}{(4\times4)+(5\times5)+(0\times0)}\\ \Rightarrow a_{1}=\frac{20+15+0}{16+25+0}\\\\ \Rightarrow a_{1}=\frac{35}{41}$

$Using\:the \:formula\: a_{2}= \frac{\textbf{y}\cdot \textbf{u}_{2}}{\textbf{u}_{2}\cdot \textbf{u}_{2}}\\ \Rightarrow a_{2}=\frac{\textbf{y}\cdot \textbf{u}_{2}}{\textbf{u}_{2}\cdot \textbf{u}_{2}}\\ \Rightarrow a_{2}=\frac{(5\times-5)+(3\times4)+(-1\times0)}{(-5\times-5)+(4\times4)+(0\times0)}\\ \Rightarrow a_{2}=\frac{-25+12+0}{25+16+0}\\\\ \Rightarrow a_{2}=\frac{-13}{41}$

Thus,

$\textbf{u} = a_{1}\textbf{u}_{1}+a_{2}\textbf{u}_{2} \:is\:the\:orthogonal\:projection\: of\:\textbf{y}\: onto\:Span\left \{ \textbf{u}_{1},\textbf{u}_{2} \right \}.\\ Thus,\\\textbf{u} =\left ( \frac{35}{41} \right )\begin{bmatrix} 4\\ 5\\ 0\\ \end{bmatrix}+\left ( \frac{-13}{41} \right )\begin{bmatrix} -5\\ 4\\ 0\\ \end{bmatrix}\\\\\\ \Rightarrow \textbf{u} =\begin{bmatrix} \frac{140}{41}\\ \\ \frac{175}{41}\\\\ 0\\ \end{bmatrix}+\begin{bmatrix} \frac{65}{41}\\ \\ \frac{-52}{41}\\\\ 0\\ \end{bmatrix}\\\\\\ \Rightarrow \textbf{u} =\begin{bmatrix} \frac{205}{41}\\ \\ \frac{123}{41}\\\\ 0\\ \end{bmatrix}\\\\\\ \Rightarrow \textbf{u} =\begin{bmatrix} 5\\ 3\\ 0\\ \end{bmatrix}$

Thus, the projection of y onto $Span\left \{ \textbf{u}_{1},\textbf{u}_{2} \right \}$ = $\begin{bmatrix} 5\\ 3\\ 0\\ \end{bmatrix}$ .

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