I am self studying Linear Algebra from Hoffman Kunze and I have a question in a theorem of Lesson-8 of text book.
Adding it's image:
How does it follows from Theorem 4 that E(c$\alpha + \beta) = cE\alpha + E\beta$ ?
Image of Statement of Theorem 4:
Kindly tell.
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 $c(E\alpha -\alpha) + (E\beta - \beta) \perp W$.