Let V and W be vector spaces, and let \(\displaystyle{f}:{V}\rightarrow{W}{f}:{V}\rightarrow{W}\) be a function from V to W. (That is, for each vector v\inV, f(v) denotes exactly one vector of W.) Then f is a linear transformation if and only if

\(\displaystyle{f{{\left({v}{1}+{c}{v}{2}\right)}}}={f{{\left({v}{1}\right)}}}+{c}{f{{\left({v}{2}\right)}}},\forall{v}{1},{v}{2}\in{V},{c}\in{R}.\)

Let V and W be vector spaces, and let \(L:V\rightarrow W\) be a linear transformation. Let 0V be the zero vector in V and 0W be the zero vector in W. Then L(0V)=0W. \(\displaystyle{\left({1}\right)}{H}{e}{r}{e}{f{{\left({x}{1},{x}{2}\right)}}}={\left({2}{x}{1}−{x}{2},{3}{x}{1}+{x}{2}\right)}{f{{\left({x}\right)}}}.\)

Then for \((x1,x2),(y1,y2)\in R^{2}(x)\) \(\displaystyle{f{{\left({\left({x}{1},{x}{2}\right)}+{c}{\left({y}{1},{y}{2}\right)}\right)}}}={f{{\left({x}{1}+{c}{y}{1},{x}{2}+{c}{y}{2}\right)}}}={\left({2}{\left({x}{1}+{c}{y}{1}\right)}−{\left({x}{2}+{c}{y}{2}\right)},{3}{\left({x}{1}+{c}{y}{1}\right)}+{\left({x}{2}+{c}{y}{2}\right)}\right)}={\left({2}{x}{1}−{x}{2}+{c}{\left({2}{y}{1}−{y}{2}\right)},{3}{x}{1}+{x}{2}+{c}{\left({3}{y}{1}+{y}{2}\right)}\right)}={\left({2}{x}{1}−{x}{2},{3}{x}{1}+{x}{2}\right)}+{c}{\left({2}{y}{1}−{y}{2},{3}{y}{1}+{y}{2}\right)}{)}={f{{\left({x}{1},{x}{2}\right)}}}+{c}{f{{\left({y}{1},{y}{2}\right)}}}.\)

Therefore ff is a linear transformation.

\(\displaystyle{\left({2}\right)}{H}{e}{r}{e}{L}{\left({x},{y},{z}\right)}={\left({x}+{y},{y}+{z},{z}+{5}\right)}{L}{\left({x},{y},{z}\right)}={\left({x}+{y},{y}+{z},{z}+{5}\right)}.\) Since \(\displaystyle{f{{\left({0},{0},{0}\right)}}}={\left({0},{0},{5}\right)}{f{{\left({0},{0},{0}\right)}}}={\left({0},{0},{5}\right)}\) not equal to (0,0,0)(0,0,0), therefore LL is not a linear transformation.

(3, 4) Try yourself using the definition.

(5) Wheather \(\displaystyle{f{{\left({2},{2}\right)}}}={f{{\left({\left({1},{1}\right)}+{\left({1},{1}\right)}={f{{\left({1},{1}\right)}}}+{f{{\left({1},{1}\right)}}}?{f{{\left({2},{2}\right)}}}={f{{\left({\left({1},{1}\right)}+{\left({1},{1}\right)}={f{{\left({1},{1}\right)}}}+{f{{\left({1},{1}\right)}}}?\right.}}}\right.}}}\)

(6) Try yourself using the definition.