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I am self studying Linear Algebra from Hoffman Kunze and I have a question in a theorem of Lesson-8 of text book.

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How does it follows from Theorem 4 that E(c$\alpha + \beta) = cE\alpha + E\beta$ ?

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Kindly tell.

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We have \begin{align} E(c\alpha + \beta) = cE\alpha + E\beta &\iff cE\alpha + E\beta \text{ is the best approximation in $W$ of $c\alpha + \beta$}\\ &\stackrel{(i)}\iff (cE\alpha + E\beta) - (c\alpha + \beta) \perp W\\ &\iff c(E\alpha -\alpha) + (E\beta - \beta) \perp W \end{align} and the last statement is true. Namely, $E\alpha$ is the best approximation in $W$ of $\alpha$ so by $(i)$ we get $E\alpha -\alpha \perp W$. Similarly $E\beta - \beta \perp W$. Therefore $c(E\alpha -\alpha) + (E\beta - \beta) \perp W$.

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  • $\begingroup$ how does E(c$\alpha+ \beta$) implies cE($\alpha$) +E($\beta$) is best approximation of W in c $\alpha$ + $\beta$ ? Can you please prove it. $\endgroup$
    – user775699
    Jul 17, 2020 at 11:09
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    $\begingroup$ @User $E(c\alpha + \beta)$ is by definition the best approximation of $c\alpha + \beta$ by vectors in $W$. Therefore $E(c\alpha + \beta) = cE\alpha + E\beta$ means that $cE\alpha + E\beta$ is the best approximation of $c\alpha + \beta$ by vectors in $W$. $\endgroup$ Jul 18, 2020 at 14:06

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