$ 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
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 $y-operatornameproj_V(y).$

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).

answer Mar 27 "16 at 20:32

Antonio Gabriele PalermoAntonio Gabriele Palermo
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