is lenear transformation given by with respect to basis
find ker T and ImT ?
How i tried to solve it:
so i got the image of three vectors.
how to find the Im T and KerT
You are given the matrix form of T and can determin image and kernel from that.
A vector, , is in the image of T if and only if there exist a vector, such that
Of course, you can solve that equation by row-reducing the augmented matrix
The vector is in the image of T if and only if row reduction does not result in an "impossible row"- a row where the first three numbers are 0 and the fourth is not.
A vector is in the kernel of T if and only if
which you can determine by row-reducing the augmented matrix
For this particular problem the easy way to do it is to look at the determinant! If the determinant is non-zero, then the matrix is invertible and so maps one to one onto . Its image is all of and its kernel is {0}.
transgalactic, i'm not sure what you mean by "find im(T)" (not that i don't know what it is, but rather what counts as "finding it").
trivially, one can say .
but even more to the point, there may be several equivalent ways of describing im(T) more explicitly.
what is usually more helpful, is finding dim(im(T)) = rank(A), and exhibiting a basis for im(T).
similarly, one finds dim(ker(T)) = nullity(A), and exhibits a basis for ker(T).
bases are in general not unique, but the dimensions of im(T) and ker(T) are.
now, if we row-reduce the matrix A, we get:
this tells us the first two columns of A form a basis for im(T), so im(T) is their span.
more explicitly, .
how did i get this? it is just
.
the rank-nullity theorem tells us that 3 = dim(im(T)) + dim(ker(T)) = 2 + dim(ker(T)), so the kernel of T has dimension 1.
so all we need to do is find one non-zero vector in ker(T), to have a basis.
applying the rref(A) matrix to and setting the result = 0
gives us a - c/2 = 0, and b + c = 0. that is: a = c/2, b= -c, so ker(T) is all elements of the form
. we can, if we wish, for convenience choose c = 2, so that a basis for ker(T) is ,
and since ,
{3x-1} is a basis for ker(T).