How do you show the direct sum of the image and kernel of the linear operator is the vector space?

Answer 1

Ok, linear algebra isn't my strongest area but I'll have a go (please note that all vectors are in bold, if a mathematical lower case letter is not in bold, assume it to be a scalar unless otherwise indicated).

First, I'm going to assume that you already know how to prove that the kernel and image of a linear operator, say f:U->V where U and V are vector spaces, is a vector subspace of U as this needs to be proved before going on to show that the sum of ker(f) and Im(f) are a vector space.

Now, a vector space must follow the following axioms (here A represents the addition axioms, and M the multiplication axioms):

A1) associativity u+(v+w)=(u+v)+w

A2) commutativity u+v=v+u

A3) identity there exists an element 0 in U such that v+0=v

A4) existence of an inverse for all u in U there exists an element -u in U such that u+(-u)=0

M1) scalar multiplication with respect to vector addition a(u+v)=au+av

M2) scalar multiplication with respect to field addition (a+b)u=au+bu

M3) compatibility of scalar and field multiplication (ab)u=a(bu)

M4) identity 1u=u where 1 is the multiplicative identity

ker(f)={u in U | f(u)=0} (or more strictly, the set of vectors u in U such that Au=0 i.e. ker(f) is the null space of the matrix A), Im(f)={v in V | f(u)=v for some u in U}

Let ker(f)+Im(f) = W. Show W is a vector space i.e. show W satisfies the above axioms.

A1, A2, M1, M2, and M3 are all trivial/simple to prove.

A3: since Im(f) is the set off all v in V obtained from f(u), assume Im(f) = V and as V is a vector space it must have, by definition, an additive identity 0, therefore W also contains 0.

A4: since ker(f) is the null space of the matrix A, f(0)=0. Now assume that f(u)=0 and f(-u)=v, since U is a vector space u+(-u)=0 so f(u+(-u))=f(0)=0 but f(u+(-u))=f(u)+f(-u)=0+vtherefore, 0=0+v which implies v=0, so if u goes to 0 in V, its inverse -u also goes to 0. A similar process for the image shows that if u goes to v then -u goes to -v. (Thanks to Jokes Free4Me for the help with this axiom)

M4: similar to A3 except V, again by definition, must have the multiplicative identity 1 and so also exists in W.

All the axioms are satisfied, therefore W=ker(f)+Im(f) is a vector space. Q.E.D.