# A question in Theorem 5 of Lesson 8 of Hoffman Kunze linear algebra

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

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 $$c(E\alpha -\alpha) + (E\beta - \beta) \perp W$$.
• 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.
• @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 $W$. Jul 18, 2020 at 14:06