9. Linear maps

Part 1.: Problems, solutions.
Part 2.: Problems, solutions.
Part 3.: Problems, solutions.

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 scalars 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).

In the above example we can also come to the conclusion what is \varphi((1,0)) and \varphi((0,1)) by writing down a matrix consisting of the vectors of the given basis on the left hand side and values on the right hand side. Each operation on the rows of the matrix does not change the principle that the vector is on the left, and its value on the right, because \varphi is a linear map. So after transforming the matrix to the reduced echelon form on the right hand side one will find \varphi((1,0)) and \varphi((0,1)).

    \[\left[\begin{array}{cc|ccc}1&-1&1&2&1\\-1&0&-1&0&1\end{array}\right]\underrightarrow{w_1\leftrightarrow w_2}\left[\begin{array}{cc|ccc}-1&0&-1&0&1\\1&-1&1&2&1\end{array}\right]\underrightarrow{w_2+w_1}\]

    \[\left[\begin{array}{cc|ccc}-1&0&-1&0&1\\0&-1&0&2&2\end{array}\right]\underrightarrow{w_1\cdot (-1),w_2\cdot (-1)}\left[\begin{array}{cc|ccc}1&0&1&0&-1\\0&1&0&-2&-2\end{array}\right],\]

so as before


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).

It means that the set of all linear maps between fixed vector spaces V and W is a vector space itself. It is denoted by L(V,W).

Composing maps

Given two linear maps \phi\colon V\to W and \psi\colon W\to U we can compose them and consider a map which transforms a vector from V first via \phi and the result via \psi getting a vector form U.

Such a map is denoted as \psi\circ \phi. Given formulas defining \phi i \psi we can easily get the formula for \psi\circ \phi. E.g. consider \phi as above and \psi\colon\mathbb{R}^2\to\mathbb{R}^4 such that \psi((a,b))=(-b,a+b,-2a-b,3a). Therefore (\psi\circ\phi)((x,y,z))=\psi(\phi((x,y,z))=\psi(((x-z,2x+3y-z))=(-2x-3y+z,3x+3y-2z,-4x-3y+3z,3x-3z).

Simple classes of linear maps

If X=V\oplus W is a direct sum of its subspaces, then \varphi\colon X\to X, \varphi(x)=v, where x=v+w, v\in V, w\in W, we call the projection onto V along W, and \psi\colon X\to X such that \varphi(x)=v-w is the reflection across V along W.

The mapping \varphi\colon W\to V, where W is a subspace of V and \varphi(x)=x is the inclusion of W into V.

For a\in K, the mapping \varphi\colon V\to V, such that \varphi(v)=av is the homothety with ratio a. For ratio 1 it is the identity, denoted by id. For scale a=0 it is the zero transformation.

Finally, the rotation by angle \theta\in \mathbb{R} is \varphi\colon \mathbb{R}^2\to\mathbb{R}^2 such that

    \[\varphi(x,y)=(x\cos\theta-y\sin\theta, x\sin \theta+y\cos\theta).\]

Kernel and image of a linear transformation

If \varphi\colon V\to W is a linear transformation, then it is easy to see that

    \[\ker\varphi=\{v\in V\colon \varphi(v)=0\}\]

called the kernel of \varphi and

    \[\text{im}\varphi=\{\varphi(v)\colon v\in V\}\]

called the image of \varphi are linear subspaces of V and W respectively. The dimension of the image of \varphi is also called the rank of \varphi and denoted by r(\varphi). Notice that if v_1,\ldots, v_n spans V, then \varphi(v_1),\ldots,\varphi(v_n) spans \text{im}\varphi. Moreover, if V=\ker\varphi\oplus U, and v_1,\ldots, v_n is a basis of U, then \varphi(v_1),\ldots,\varphi(v_n) is a basis of \text{im}\varphi. Therefore,

    \[\dim V=\dim\ker\varphi+\dim\text{im}\varphi.\]

Monomorphisms, epimorphisms and isomorphisms

A linear map which is one-to-one is called a monomorphism. It a map is ,,onto” it is called an epimorphism. If it is a bijection, we call it isomorphism.

E.g. if V=V_1\oplus V_2 then the projection \varphi\colon V\to V onto V_1 along V_2 is neither epi nor mono, but if considered as a map \varphi\colon V\to V_1 it is an epimorphism. The reflection across V_1 along V_2 is an isomorphism. Inclusion W\to V of a subspace W into V is an example of a monomorphism.

It is easy to see that a map \varphi\colon V\to W is a monomorphism if and only if \ker \varphi =\{0\}, but it is an epimorphism if \text{im}\varphi = W.

Therefore, every monomorphism maps every linearly independent system of vectors onto a linearly independent system. Thus, every isomorphism maps every basis onto a basis. Indeed, a mapping is an isomorphis if and only if it maps a basis onto a basis. Therefore, if \varphi\colon V\to W is a isomorphism, then \dim V=\dim W. Moreover, if V is finite and \dim V=\dim W, then the notions of isomorphism, monomorphism and epimorphism V\to W coincide.

Two spaces are isomorphic if there is an isomorphism between them, and we denote this by V\simeq W. Notice that two finitely dimensional subspaces over a field K are isomorphic if and only if the dimensions are equal (to see that it is true, consider bases!).

Inverse mappings

It is easy to prove that \varphi\colon V\to W is

  • an epimorphism if and only if there exists a mapping \psi\colon W\to V such that \varphi\circ\psi=id,
  • a monomorphism if and only if there exists a mapping \psi\colon W\to V such that \psi\circ\varphi=id,
  • an isomorphism if and only if there exists a mapping \psi\colon W\to V such that \varphi\circ\psi=id and \psi\circ\varphi=id.

In the last case \psi is called the inverse map to \varphi and denoted by \varphi^{-1}.

E.g. if \varphi\colon\mathbb{R}^2\to\mathbb{R}^2 and \varphi((x,y))=(x+y,x), then \varphi^{-1}((a,b))=(b,a-b).