# Another question in linear algebra.

1. Jun 24, 2007

### MathematicalPhysicist

I need to prove that if T1 and T2 are linear operators, then
dim(kerT1)+dim(KerT2)>=dim(Ker(T1oT2)).

now, i thought to show that Ker(T1oT2) is a subset of ker(T1), and then it obviously follows, but here im stuck: if u in ker(T1oT2) then T1(T2(u))=0 so T2(u) is in Ker(T1) but i need to show that u is in kerT1, if it's correct.

any hints.

2. Jun 24, 2007

### matt grime

Can I change it to S and T, rather than T1 and T2, which, let's face it, is a very bad notation.

ker(ST) is certainly not a subset (or even subspace) of ker(S). There is no reason to suppose that T maps ker(S) into ker(S), which is what you are trying (and therefore failing) to show.

u is in ker(ST) if either T(u)=0, or T(u) lies in ker(S). Of course T(u)=0 implies that T(u) is in ker(S), thus all you need to do is to work out the dimension of the preimage of ker(S) under T.

The pre-image of any subspace V under a map T is a subspace of the form U+ker(T), with the sum not necessarily direct, and T mapping U isomorphically to V. (This is just the isomorphism theorems in action.) This completely solves your problem, and was just writing out the definitions.

Last edited: Jun 24, 2007
3. Jun 24, 2007

### jambaugh

You have things reversed.
$$\mathop{ker}(T_2) \subseteq \mathop{ker}(T_1\circ T_2)$$

But that won't prove your inequality either but it is one step. Also consider the space
$$\mathbf{N}= \mathop{ker}(T_1\circ T_2) - \mathop{ker}(T_2)$$
and its image:
$$\mathbf{N}'= T_2(\mathbf{N})$$

Last edited: Jun 24, 2007
4. Jun 25, 2007

### MathematicalPhysicist

so you mean i need to find the dimension of T^-1(kerS), right?

5. Jun 25, 2007

### MathematicalPhysicist

ah, ok i see it's only an application of dim(ImT)+dim(KerT)=dimV.

6. Jun 25, 2007

There ya go.