\$ u_1=eginbmatrix2 \5\-1 endbmatrix, u_2=eginbmatrix-2 \1\1 endbmatrix,y=eginbmatrix1 \2\3 endbmatrix\$

I"m searching for the projection of \$y\$ top top \$V^perp\$, discovering that \$V\$ is created by \$S=u_1,u_2\$.*

I walk the estimate of \$proj_V(y)\$:eginalign*proj_V(y)&=fracu_1+frac^2u_2\&=frac310u_1+frac12u_2endalign*

But ns don"t know how to find the projection of \$y\$ on \$V^perp\$.Do you have any kind of hint? i don"t perefectly recognize what \$V^perp\$ is...

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linear-algebra orthogonality
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request Nov 16 "15 at 13:45

Revolucion because that MonicaRevolucion because that Monica
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The vector \$operatornameproj_V(y)\$ is the facet of \$V\$ the is closest come \$y.\$ \$V^perp\$ is the set of all vectors \$v\$ such that \$langle u,v angle=0\$ for every \$uin V\$, or (geometrically) the collection of vectors that room orthogonal to every \$V\$-vector. Now, take into consideration \$y-operatornameproj_V(y).\$

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reply Nov 16 "15 at 13:55

Cameron BuieCameron Buie
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You can additionally simply uncover a (nonzero) vector in \$V ^ot\$ (how?), and also then calculation the orthogonal forecast of y onto this vector. Exercise:prove that the two techniques are equivalent(you have the right to think come the orthogonal decomposition of a vector top top a offered direction).

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answer Mar 27 "16 at 20:32

Antonio Gabriele PalermoAntonio Gabriele Palermo
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