# direct sum

• Dec 16th 2013, 02:03 PM
sipnayan
direct sum
show that if f is in Hom(U,U) such that f2=f then V= ker f + ker(f - idv). also let B0 be an ordered basis For ker f and B1 be ordered basis for ker(f-idv). if B is an ordered basis and B=B0 + B1 for V. what is the matrix representation of f given B,B?
• Dec 16th 2013, 02:21 PM
SlipEternal
Re: direct sum
Where are you having trouble?
• Dec 16th 2013, 02:25 PM
sipnayan
Re: direct sum
in everything. i dont understand sir
• Dec 16th 2013, 03:00 PM
SlipEternal
Re: direct sum
Are you looking for a place to start? Let's start with the first statement. How do you show that two sets are equal?
• Dec 16th 2013, 03:18 PM
sipnayan
Re: direct sum
subsets of each other
• Dec 16th 2013, 03:19 PM
sipnayan
Re: direct sum
i dont know what it means if it is in kerf and ker (f-idv)
• Dec 16th 2013, 03:25 PM
SlipEternal
Re: direct sum
Do you know what the kernel of a homomorphism is? If v is in the kernel of f, then f(v) = 0. If v is in the kernel of (f-id_V), then (f-id_V)(v)=0. So, the first thing you want to do is show that V=ker f (+) ker (f-id_V) is a subspace of U (where (+) is direct sum since I am having trouble with LaTeX).
• Dec 16th 2013, 03:31 PM
sipnayan
Re: direct sum
its ok. how to show that it is direct sum?
• Dec 16th 2013, 03:35 PM
SlipEternal
Re: direct sum
Show that ker f intersect ker (f-id_V) is only the zero vector.
• Dec 16th 2013, 03:36 PM
sipnayan
Re: direct sum
and how should i do it?
• Dec 16th 2013, 03:41 PM
SlipEternal
Re: direct sum
I just told you. There is nothing magic about it. Suppose v is in ker f and v is in ker(f-id_V). Show that v is the zero vector. Since v is in ker f, you have f(v)=0. Since v is in ker(f-id_V), (f-id_V)(v) = 0. For the second one, you have (f-id_V)(v) = f(v) - id_V(v) = f(v)-v = 0. If f(v)-v=0 then f(v)=v. If f(v)=v and f(v)=0 then by transitivity, v=0.
• Dec 16th 2013, 03:46 PM
sipnayan
Re: direct sum
thanks a lot. how about the matrix representation?
• Jan 3rd 2014, 03:36 PM
Deveno
Re: direct sum
Let v be any element of V.

Then if w = f(v), we have:

f(w) = f(f(v)) = f2(v) = f(v) = w.

Thus f(w) - w = 0, so:

(f - idV)(w) = 0, that is: f(v) is in ker(f - idV).

Write:

v = f(v) + v - f(v) (which is always trivially true).

We have, by the above, that f(v) is in ker(f - idV).

Now let's compute f(v - f(v)):

f(v - f(v)) = f(v) - f(f(v)) = f(v) - f2(v) = f(v) - f(v) = 0, so for ANY v, v - f(v) is in ker(f).

So this tells us that we can write any element of V as the sum of something in ker(f) (namely: v - f(v)) and something in ker(f - idV) (namely: f(v)).

Hence V = ker(f) + ker(f - idV).

Now SlipEternal has already shown that the intersection of these two sets is {0}, hence the sum is direct.

As for the matrix representation, we have:

[f]B = [0 0...0 b1 b2....bk] (these are column vectors) where B1 = {b1,...,bk}

Here is a simple example:

Let V = R2, and let f(x,y) = (x,0).

Then f2(x,y) = f(f(x,y)) = f(x,0) = (x,0), so we see that f2 = f.

Clearly, a basis for ker(f) is {(0,1)}. let's find a basis for ker(f - idV).

If f(x,y) = (x,y), then:

(x,0) = (x,y), so y = 0. Thus a basis is: {(1,0)}. In the ordered basis B = {(0,1),(1,0)} (note this is not the USUAL ordered basis), we have

[f]B =

[0 1]
[0 0].

(Slight note: usually we write the ordered basis the "other way around" as B = B1 + B0, so the "0's come last").