[gwnn]

Sorry to take up everyone's time... But I know most people here are mathematically inclined and as far as I've seen, very very helpful.. Tomorrow I have this algebra exam and while I think I got the hang of most of it, this (sort of a) problem seems to remain elusive (for me):

Let f be an endomorphism of vector space R R^4 given by

(f)e= this matrix with the lines

(0 1 2 3)
(-1 2 1 0)
(3 0 -1 -2)
(5 -3 -1 1)

in canonical basis.

Determine the dimension and give a basis for each of the following:

Im f; Ker f; Im f+Ker f; Im f (intersected with) Ker f.

I have some of the notes but only for the Ker f thingie, which is quite trivial from the definition that the kernel is every vector x for which f(x)=0 vector. But from there I don't really understand... Im f is every possible result of the linear transformation? But then Imf+Kerf is basically Imf? Or what?

Can anyone pls help me? Thanks very very much!

[Trumpace]

Do you have a textbook for this? I think you should look that up for the followed notation.

Anyway, Im f, if stands for image is the set of vectors y, such that there is a vector x for which f(x) = y.


The usual definition for A+B i have seen is: if A and B are two vector spaces, then A + B = { a + b | a \in A, b \in B }

It could also be Union of Im f and Ker f, but in that case it might not make much sense to talk about a basis, as it might not be a vector space.

Hope that helps.
Good luck with your exam.

[david_c]

In this example, a basis for Im f is

( 1 )     ( 0 )     ( 0 )
( 0 )     ( 1 )     ( 0 )
( 0 )  ,  ( 0 )  ,  ( 1 )
( 1 )     (-2 )     ( 1 )

(Of course, there are lots of other ways of writing this. Choosing any three columns of the original matrix would work, but it's better to get a more reduced form.)

And a basis for Ker f is

( 0 )
( 1 )
(-2 )
( 1 )

I assume you know how to work out these things.

Im f + Ker f is then the space spanned by all four vectors. Since the four vectors are clearly linearly independent, Im f + Ker f has dimension 4, ie. it's the whole of R^4.

Im f (intersected with) Ker f is {0} (once you have a nice basis for Im f this is easy to see).

If the four vectors weren't linearly independent then Im f + Ker f would have dimension 3; but it would have to contain all of Im f, so it would have to be exactly the same as Im f. In this case, Im f (intersected with) Ker f would be the whole of Ker f.

[Hope the working is correct...]