6. Linear maps

A linear map \phi is a map which maps vectors from a given space V, to vectors from another linear space W (\phi\colon V\to W), and satisfying the linear condition, which says that for every vectors v,v'\in V and numbers a,b we have \phi(av+bv')=a\phi(v)+b\phi(v'). E.g. a rotation around (0,0) is a linear map \mathbb{R}^2\to\mathbb{R}^2. Given two vectors and digits we will get the same vector regardless of whether we rotate the vector first and then multiply by numbers and add them, or multiply by numbers, add and then rotate.

Therefore, to prove that a given map is a linear map we need to prove that for any two vectors and any two numbers it satisfies the linear condition. E.g. \varphi\colon \mathbb{R}^2\to\mathbb{R} given as \varphi(x,y)=-x+2y is a linear map because if (x,y),(x',y')\in\mathbb{R}^2 and a,b\in\mathbb{R}, then \varphi(a(x,y)+b(x',y'))=\varphi((ax+bx',ay+by'))=-ax-bx'+2ay+2by'=a(-x+2y)+b(-x'+2y')=a\varphi((x,y))+b\varphi((x',y')).

Meanwhile, to disprove that a map is linear we need to find an example of two vectors along with two numbers such that the linear condition fail for them. E.g. \psi\colon \mathbb{R}\to\mathbb{R}^2 given as \psi(x)=(2x+1,0) is not a linear map because \psi(1+1)=\psi(2)=(5,0)\neq (6,0)=(3,0)+(3,0)=\psi(1)+\psi(1).

Usually, linear maps will be given by their formulas. E.g. \psi\colon\mathbb{R}^2\to\mathbb{R}^2, \psi((x,y))=(y,-x). Then to see what \psi does to a given vector, e.g. (1,2), we substitute it to the formula: \psi((1,2))=(2,-1). By the way, it is easy to see, that this is simply the rotation around (0,0) by 90 degrees clockwise. Less geometrical example: let \phi\colon\mathbb{R}^4\to\mathbb{R}^2, \phi((x,y,z,t))=(x+3y-t,3y-2z+t), therefore \phi maps vector (1,2,0,-1) to vector (1+6-(-1),6-0+(-1))=(8,5).

Sometimes we can define a linear map by giving its values on the vectors from a given basis only. This suffices to determine this map. E.g. let \varphi\colon\mathbb{R}^2\to\mathbb{R}^3 be given in the following form: \varphi((1,-1))=(1,2,1), \varphi((-1,0))=(-1,0,1) (vectors (1,-1), (-1,0) constitute a basis of the plane). We can calculate the formula of this map. First calculate the coefficients of the standard basis in the given one (by solving a system of equations or by guessing). In this case we see that (1,0)=0\cdot(1,-1)-1(-1,0) (coefficients: 0,-1 and (0,1)=-(1,-1)-(-1,0) (coefficients: -1,-1). Therefore \varphi((x,y))=x\varphi((1,0))+y\varphi((0,1))=x(-\varphi((-1,0)))+y(-\varphi((1,-1))-\varphi((-1,0)))=x(-(-1,0,1))+y(-(1,2,1)-(-1,0,1))=x(1,0,-1)+y(0,-2,-2)=(x,-2y,-x-2y).

Given two linear maps \varphi,\psi\colon V\to W and a number a, their sum \varphi+\psi and a\varphi are linear maps V\to W. Obviously we add and multiply by coefficients. So e.g. if \varphi((x,y))=(-x,-2y),\psi((x,y))=(0,2x), then (\varphi+\psi)((x,y))=(-x,2x-2y).