Question

# For each problem below, either prove that the mapping is linear or explain why it cannot be linear. 1. f(x_1,x_2 )=(2x_1-x_2,3x_1+x_2) 2. L(x,y,z)=(x+y,y+z,z+5) 3. L(x,y)=(x+y,0,x-2y) 4. f(x,y)=(2x+y,-3x+5y) 5. f(x,y)=(x^2,x+y) 6. L(x,y)=(x,x+y,-y)

Vectors and spaces
For each problem below, either prove that the mapping is linear or explain why it cannot be linear.
$$\displaystyle{1}.{f{{\left({x}_{{1}},{x}_{{2}}\right)}}}={\left({2}{x}_{{1}}-{x}_{{2}},{3}{x}_{{1}}+{x}_{{2}}\right)}$$
$$\displaystyle{2}.{L}{\left({x},{y},{z}\right)}={\left({x}+{y},{y}+{z},{z}+{5}\right)}$$
$$\displaystyle{3}.{L}{\left({x},{y}\right)}={\left({x}+{y},{0},{x}-{2}{y}\right)}$$
$$\displaystyle{4}.{f{{\left({x},{y}\right)}}}={\left({2}{x}+{y},-{3}{x}+{5}{y}\right)}$$
$$\displaystyle{5}.{f{{\left({x},{y}\right)}}}={\left({x}^{{2}},{x}+{y}\right)}$$
$$\displaystyle{6}.{L}{\left({x},{y}\right)}={\left({x},{x}+{y},-{y}\right)}$$

2021-01-06

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.