# 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)

For each problem below, either prove that the mapping is linear or explain why it cannot be linear.
$1.f\left({x}_{1},{x}_{2}\right)=\left(2{x}_{1}-{x}_{2},3{x}_{1}+{x}_{2}\right)$
$2.L\left(x,y,z\right)=\left(x+y,y+z,z+5\right)$
$3.L\left(x,y\right)=\left(x+y,0,x-2y\right)$
$4.f\left(x,y\right)=\left(2x+y,-3x+5y\right)$
$5.f\left(x,y\right)=\left({x}^{2},x+y\right)$
$6.L\left(x,y\right)=\left(x,x+y,-y\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

lobeflepnoumni

Let V and W be vector spaces, and let $f:V\to Wf:V\to W$ 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
$f\left(v1+cv2\right)=f\left(v1\right)+cf\left(v2\right),\mathrm{\forall }v1,v2\in V,c\in R.$
Let V and W be vector spaces, and let $L:V\to W$ be a linear transformation. Let 0V be the zero vector in V and 0W be the zero vector in W. Then L(0V)=0W.​ $\left(1\right)Heref\left(x1,x2\right)=\left(2x1-x2,3x1+x2\right)f\left(x\right).$
Then for $\left(x1,x2\right),\left(y1,y2\right)\in {R}^{2}\left(x\right)$ $f\left(\left(x1,x2\right)+c\left(y1,y2\right)\right)=f\left(x1+cy1,x2+cy2\right)=\left(2\left(x1+cy1\right)-\left(x2+cy2\right),3\left(x1+cy1\right)+\left(x2+cy2\right)\right)=\left(2x1-x2+c\left(2y1-y2\right),3x1+x2+c\left(3y1+y2\right)\right)=\left(2x1-x2,3x1+x2\right)+c\left(2y1-y2,3y1+y2\right)\right)=f\left(x1,x2\right)+cf\left(y1,y2\right).$
Therefore ff is a linear transformation.
$\left(2\right)HereL\left(x,y,z\right)=\left(x+y,y+z,z+5\right)L\left(x,y,z\right)=\left(x+y,y+z,z+5\right).$ Since $f\left(0,0,0\right)=\left(0,0,5\right)f\left(0,0,0\right)=\left(0,0,5\right)$ 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 $f\left(2,2\right)=f\left(\left(1,1\right)+\left(1,1\right)=f\left(1,1\right)+f\left(1,1\right)?f\left(2,2\right)=f\left(\left(1,1\right)+\left(1,1\right)=f\left(1,1\right)+f\left(1,1\right)?$
(6) Try yourself using the definition.