Thread: Linear maps

Consider the linear map defined on with values in given by the matrix:
A =
( 1 3 1 -2 )
( 3 9 4 -4 ) yes, this is supposed to be a matrix, i'm new
(-1 -3 -2 0 )
a) find all x e satisfying the linear equation Ax=b, with
b =
( 1)
(-2)
( 4)

b) find a basis for ker A

c) Using the Rank Theorem, find the dimension of im A

This is a past paper without solutions and would really help me revise, thank you very much !

Well, I don't know if there is an easier way than this, but I suspect
there is:

a) If you multiply Ax=b then you will get three equations for which
3 of the variables in x can be eliminated leaving you with




so


b) the kernel of A can be found in a similar way Ax=0




which is a straight line through the origin,

so a basis could be which has dimension 1, ie n(A)=1

c) I'm not sure what the rank theorem is but r(A)+n(A)=dimX means
that r(A)+1=4 so n(A)=3. But I thought that was kind of obvious anyway,
so maybe I misunderstood the question.

The "rank theorem" does, indeed, say that, if A is a linear operator from vector space U to vector space V, then the rank of A plus the nullity (dimension of the kernel) of A is equal to the dimension of U. And you have a typo:
r(A)+ n(A)= 4 and n(A)= 1 so r(A)= 3, not "n(A)= 3". That says that A maps onto which was not entirely "obvious". The rank of A, the dimension of its image in must be less than or equal to 3 but certainly might have been less.

Ah, I see, thanks for clearing that one up HallsofIvy.