Exercise 13Linear algebra › Linear mapsHigh schoolAdded 3 Nov 2026

Checking linearity

#linear map#definition#examples

Statement

Determine whether each map is linear:

(a) $f:\mathbb{R}^2\to\mathbb{R}^2$ , $f(x,y)=(2x-y,\ x+3y)$ ;
(b) $g:\mathbb{R}^2\to\mathbb{R}$ , $g(x,y)=xy$ ;
(c) $h:\mathbb{R}\to\mathbb{R}$ , $h(x)=x+1$ ;
(d) $k:\mathbb{R}^2\to\mathbb{R}^2$ , $k(x,y)=(x^2,\,y)$ .

Hint

A map $\varphi$ is linear if $\varphi(u+v)=\varphi(u)+\varphi(v)$ and $\varphi(\lambda u)=\lambda\varphi(u)$ for all vectors $u,v$ and scalars $\lambda$ . A quick necessary condition is $\varphi(0)=0$ (take $\lambda=0$ ); it is not sufficient.

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Solution

(a) For $(x,y),(x',y')\in\mathbb{R}^2$ and $\lambda\in\mathbb{R}$ :
$f(\lambda(x,y))=(2\lambda x-\lambda y,\ \lambda x+3\lambda y)=\lambda f(x,y),$
$f\big((x,y)+(x',y')\big)=\big(2(x+x')-(y+y'),\ (x+x')+3(y+y')\big)=f(x,y)+f(x',y').$
So $f$ is linear.

(b) $g(2,2)=4$ but $2g(1,1)=2$ , so $g(2\cdot(1,1))\neq 2g(1,1)$ . Not linear. (In general $g(\lambda x,\lambda y)=\lambda^2xy$ .)

(c) $h(0)=1\neq0$ , whereas a linear map satisfies $h(0)=h(0\cdot 0)=0\cdot h(0)=0$ . Not linear.

(d) $k(2,0)=(4,0)$ but $2k(1,0)=(2,0)$ , so homogeneity fails. Not linear. (Note that $k(0,0)=(0,0)$ : the condition $\varphi(0)=0$ is necessary but not sufficient. The same remark applies to (b).)

Conclusion. Only (a) is linear.

Checking linearity

Statement

Solution

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