# Show that if u is a vector in R^{2} or R&3, then u+(−1)u=0 u+(−1)u=0

Question
Alternate coordinate systems
Show that if u is a vector in R^{2} or R&3, then $$\displaystyle{u}+{\left(−{1}\right)}{u}={0}$$
$$\displaystyle{u}+{\left(−{1}\right)}{u}={0}$$

2021-01-29
First, we will show that 0⋅u=0 for any vector u∈R^{2} or R^{3}. Now ​
PSK0\times u+0\times u=(0+0)\times u=0\times u \Rightarrow 0\times u+0\times u+(−0\times u)=0\times u+(−0⋅u) \Rightarrow 0\times u+[0\times u+(−0\times u)]=0\times u+(−0⋅u) \Rightarrow 0\times u+0=0 \Rightarrow 0\times u=0.ZSK ​
Therefore ​
$$\displaystyle{u}+{\left(−{1}\right)}\times{u}={1}\times{u}+{\left(−{1}\right)}\times{u}={\left({1}+{\left(−{1}\right)}\right)}\times{u}{)}={0}\times{u}={0}.$$
PSKu+(−1)\times u =1\times u+(−1)\times u =(1+(−1))\times u) =0\times u =0.ZSK ​

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All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$\displaystyle{R}^{{2}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\$$