Exercise 4.5.17

Answers

Recall the definition

(δ_{1} + δ_{2}) (A) = δ_{1} (A) + δ_{2} (A)

and

(kδ) (A) = kδ (A)

. For brevity, we write $δ^{'}$ for $δ_{1} + δ_{2}$ and $δ^{″}$ for $kδ$ . Now prove that both $δ^{'}$ and $δ^{″}$ is $n$ -linear. Check that

δ^{'} (\begin{matrix} u_{1} + c v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) = δ_{1} (\begin{matrix} u_{1} + c v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + δ_{2} (\begin{matrix} u_{1} + c v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix})

= δ_{1} (\begin{matrix} u_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + c δ_{1} (\begin{matrix} v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + δ_{2} (\begin{matrix} u_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + δ_{2} (\begin{matrix} v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix})

= δ^{'} (\begin{matrix} u_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + c δ^{'} (\begin{matrix} v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) .

Also check that

δ^{″} (\begin{matrix} u_{1} + c v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) = kδ (\begin{matrix} u_{1} + c v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix})

= kδ (\begin{matrix} u_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + ckδ (\begin{matrix} v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) = δ^{″} (\begin{matrix} u_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) + c δ^{″} (\begin{matrix} v_{1} \\ u_{2} \\ ⋮ \\ u_{n} \end{matrix}) .

So both $δ^{'}$ and $δ^{″}$ is linear function for the first row when other rows are held fixed. For the cases on other rows are similar.

jephianlin

2011-06-27 00:00