Question

Prove part (a) of Theorem 3.4.2.

Short answer

If $\mathfrak{I}$be the basis of a subspace V of ${\mathrm{ℝ}}^n$then ${\left[\stackrel{\rightharpoonup }{x}+\stackrel{\rightharpoonup }{y}\right]}_{\mathfrak{I}}={\left[\stackrel{\rightharpoonup }{x}\right]}_{\mathfrak{I}}+{\left[\stackrel{\rightharpoonup }{x}\right]}_{\mathfrak{I}}$for all vectors $\stackrel{\rightharpoonup }{\mathrm{x}}\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{y}}$in V.

Step-by-step solution

Given

If$\mathfrak{I}$be the basis of a subspace V of ${\mathrm{ℝ}}^n$then

${\left[\stackrel{\rightharpoonup }{\mathrm{x}}+\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}+{\left[\stackrel{\rightharpoonup }{y}\right]}_{\mathfrak{I}}$ for all vectors in $\stackrel{\rightharpoonup }{\mathrm{x}}\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{y}}$V.

Finding the coordinate vectors of x⇀  and y⇀ 

Let $\mathfrak{I}=\left\{{\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_2,...,{\stackrel{\rightharpoonup }{v}}_n\right\}$be the basis of the subspace V of${\mathrm{ℝ}}^n$

Then any vectors $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in {\mathrm{ℝ}}^n$can be uniquely represented as a linear combinations of the vectors ${\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_2,...,{\stackrel{\rightharpoonup }{v}}_n$. Then we have

localid="1660645060506" $$\begin{gathered} \stackrel{\rightharpoonup }{x}=a_1{\stackrel{\rightharpoonup }{v}}_1+a_2{\stackrel{\rightharpoonup }{v}}_2+...+a_n{\stackrel{\rightharpoonup }{v}}_n \\ and \\ \stackrel{\rightharpoonup }{y}=b_1{\stackrel{\rightharpoonup }{v}}_1+b_2{\stackrel{\rightharpoonup }{v}}_2+...+b_n{\stackrel{\rightharpoonup }{v}}_n \end{gathered}$$

localid="1660645555472" $\Rightarrow {\left[\stackrel{\rightharpoonup }{x}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{a}_1\\ a_2\\ .\\ .\\ .\\ a_n\end{array}\right]\mathrm{and}{\left[\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{\mathrm{b}}_1\\ {\mathrm{b}}_2\\ .\\ .\\ .\\ {\mathrm{b}}_{\mathrm{n}}\end{array}\right]$

Finding the coordinate vectors ofx⇀+y⇀

Since we have

$$\begin{gathered} \stackrel{\rightharpoonup }{x}=a_1{\stackrel{\rightharpoonup }{v}}_1+a_2{\stackrel{\rightharpoonup }{v}}_2+...+a_n{\stackrel{\rightharpoonup }{v}}_n \\ and \\ \stackrel{\rightharpoonup }{y}=b_1{\stackrel{\rightharpoonup }{v}}_1+b_2{\stackrel{\rightharpoonup }{v}}_2+...+b_n{\stackrel{\rightharpoonup }{v}}_n \end{gathered}$$

Where

$$\begin{gathered} \Rightarrow \stackrel{\rightharpoonup }{x}+\stackrel{\rightharpoonup }{y}=a_1{\stackrel{\rightharpoonup }{v}}_1+a_2{\stackrel{\rightharpoonup }{v}}_2+...+a_n{\stackrel{\rightharpoonup }{v}}_n\stackrel{\rightharpoonup }{x}+b_1{\stackrel{\rightharpoonup }{v}}_1+b_2{\stackrel{\rightharpoonup }{v}}_2+...+b_n{\stackrel{\rightharpoonup }{v}}_n+ \\ \Rightarrow \stackrel{\rightharpoonup }{x}+\stackrel{\rightharpoonup }{y}=(a_1+b_1{\stackrel{\rightharpoonup }{)v}}_1+(a_2+b_2){\stackrel{\rightharpoonup }{v}}_2+...(a_n+b_n){\stackrel{\rightharpoonup }{v}}_n \\ \Rightarrow \stackrel{\rightharpoonup }{x}+\stackrel{\rightharpoonup }{y}=c_1{\stackrel{\rightharpoonup }{v}}_1+c_2{\stackrel{\rightharpoonup }{v}}_2+...+c_n{\stackrel{\rightharpoonup }{v}}_n \\ \mathrm{Where}{\mathrm{c}}_{\mathrm{i}}={\mathrm{a}}_{\mathrm{i}}+{\mathrm{b}}_{\mathrm{i}},\mathrm{i}=1,2,3,...\mathrm{n}. \\ \Rightarrow {\left[\stackrel{\rightharpoonup }{\mathrm{x}}+\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{\mathrm{c}}_1\\ {\mathrm{c}}_2\\ .\\ .\\ .\\ {\mathrm{c}}_{\mathrm{n}}\end{array}\right] \end{gathered}$$

  To prove [x⇀+y⇀]I=[x⇀]I+[y⇀]I

We have

$$\begin{gathered} \Rightarrow {\left[\stackrel{\rightharpoonup }{x}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{a}_1\\ a_2\\ .\\ .\\ .\\ a_n\end{array}\right]\mathrm{and}{\left[\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{\mathrm{b}}_1\\ {\mathrm{b}}_2\\ .\\ .\\ .\\ {\mathrm{b}}_{\mathrm{n}}\end{array}\right] \\ \mathrm{And} \\ \Rightarrow {\left[\stackrel{\rightharpoonup }{\mathrm{x}}+\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{\mathrm{c}}_1\\ {\mathrm{c}}_2\\ .\\ .\\ .\\ {\mathrm{c}}_{\mathrm{n}}\end{array}\right] \\ {\left[\stackrel{\rightharpoonup }{\mathrm{x}}+\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}=\left[\begin{array}{c}{\mathrm{c}}_1\\ {\mathrm{c}}_2\\ .\\ .\\ .\\ {\mathrm{c}}_{\mathrm{n}}\end{array}\right]=\left[\begin{array}{c}{a}_1\\ a_2\\ .\\ .\\ .\\ a_n\end{array}\right]+\left[\begin{array}{c}{\mathrm{b}}_1\\ {\mathrm{b}}_2\\ .\\ .\\ .\\ {\mathrm{b}}_{\mathrm{n}}\end{array}\right] \\ \Rightarrow {\left[\stackrel{\rightharpoonup }{\mathrm{x}}+\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}={\left[\stackrel{\rightharpoonup }{x}\right]}_{\mathfrak{I}}+{\left[\stackrel{\rightharpoonup }{y}\right]}_{\mathfrak{I}} \end{gathered}$$

  Final Answer

If$$\begin{gathered} \mathfrak{I}=\left\{{\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_2,...,{\stackrel{\rightharpoonup }{v}}_n\right\}\mathrm{be}\mathrm{the}\mathrm{basis}\mathrm{of}\mathrm{a}\mathrm{subspace}\mathrm{V}\mathrm{of}{\mathrm{ℝ}}^{\mathrm{n}},\mathrm{then}{\left[\stackrel{\rightharpoonup }{\mathrm{x}}+\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}={\left[\stackrel{\rightharpoonup }{\mathrm{x}}\right]}_{\mathfrak{I}}+{\left[\stackrel{\rightharpoonup }{\mathrm{y}}\right]}_{\mathfrak{I}}\mathrm{for} \\ \mathrm{all}\mathrm{vectors}\stackrel{\rightharpoonup }{\mathrm{x}}\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{y}}\mathrm{in}\mathrm{V}. \end{gathered}$$be