Wednesday 19012011, Lecture notes by Y. Burda 1 Rank-nullity theorem A very simple example of kernel and image (with picture): Let f : R R be given by f(( )) y ( ) , i.e. the orthogonal projection 0 x onto the x-axis. The kernel is by denition the set of all vectors ( y) such x x 0 that their image under f is the zero vector, i.e. {( y)( 0 = ( )0. Thus kerf = ( ). 1 x The image of f is by denition the set of all images of vectors ( y) under x x 2 x 1 f: Imf = {f ( y( )y R } = {( )0 R} = ( 0). It turns out that the rank and the nullity of a transformation are always related by a very simple formula, called rank-nullity theorem. Theorem (Rank-nullity theorem). Let f : V W be a linear transforma- tion of K-vector spaces. Assume that V is nite-dimensional. Then dimkerf + dimImf = dimV Proof. The subspace kerf is nite-dimensional, since it is a subspace of a nite-dimensional space V . Choose its basis v ,...,v . Being a basis of some 1 k subspace, the vectors v 1...,v ake linearly independent. We can complete them to a basis of V by adding some new vectors u ,..1,u . m For now we have the following: v1,...,vkis a basis of kerf and v ,1..,v ,k ,.1.,u m is a basis of V ; in particular kerf is k-dimensional, while V itself is k + m-dimensional. Since V is spanned by the vectors v ,1..,v ,uk,.1.,u , ims image f(V ) is spanned by the vectors f(v 1,...,f(v )kf(u ),1..,f(u ). motice however that f(v 1 = 0,...,f(v ) k 0 since v ,..1,v liekin the kernel of f. Thus Imf = f(u )1...,f(u )m If we can prove that f(u 1,...,f(u )mare also linearly independent, well be able to deduce that they form a basis of Imf and in particular the dimen- sion of Imf would have to be m, thus proving the theorem. So now we prove that f(u ),...,f(u ) are linearly independent: 1 m Suppose that 1(u )1...+ f(um) = m. Then f( u +...1 1 ) = 0 m m since f is linear. It means that 1 1 ... + um m is in the kernel of f. In particular it can be written as a linear combination of basis vectors of kerf: 1 1 ... + um m = 1 1 ... + vk k 1 www.notesolution.com

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