# 2 One to One Linear Algebra Proofs?

Does anyone know how to prove these?

Let T1: Rn->Rk, and T2: Rk ->Rm be linear transformations.

Prove that ker(T1) and ker(T2 * T1) are subspaces of the same Euclidean space. Is this space Rn, Rk, or Rm?

Prove that ker(T1) belongs to the subset of ker(T2 * T1)

Any help on these would be greatly appreciated. Thanks!

Relevance

Suleiman hast totally omitted the proof from his answer...

Review the definition: for T:A→B

ker(T) = { a in A : T(a) = 0 }

First notice:

T1:Rn→Rk, so ker(T1) ⊆ Rn.

T2*T1:Rn→Rm, so ker(T1) ⊆ Rn too.

Now you must prove, and we can do it in general, that any linear transformation T:A→B has a kernel that is a linear subspace of A. We know it is a subset of A, from the definition above.

We must prove the three axioms:

1) ker(T) contains 0

2) ker(T) is closed under +

3) ker(T) is closed under multiplication by scalars

STEP 1:

First of all, a linear transformation must have T(0) = 0, since it is linear. To prove this, we consider the fact that:

2T(0) = T(0) + T(0) = T(0+0) = T(0)

Subtract T(0) from both sides to get:

T(0) = 0

So 0 is in ker(T) since T(0) = 0.

STEP 2:

Now consider a,b in ker(T). That means T(a)=T(b)=0.

T(a) = 0

T(b) = 0

T(a) + T(b) = 0

But T(a)+T(b) = T(a+b) since T is linear. Thus a+b is in ker(T) too.

STEP 3:

Finally, if a is in ker(T), then T(a)=0. Multiply both sides by c:

c T(a) = c 0 = 0

But since T is linear, c T(a) = T(ca). Thus T(ca)=0, thus ca is in ker(T).

That is the proof. That is a complete answer for the first part. The second part is simple. To prove something is a subset, you follow the formula a in A implies a in B means A⊆B.

So take a in ker(T1).

Then T1(a) = 0.

Apply T2 to both sides:

T2(T1(a)) = T2(0) = 0

Thus a is also in ker(T2 * T1).

Thus ker(T1) ⊆ ker(T2 * T1).

Again, that is the complete proof.

"Prove that ker(T1) and ker(T2 * T1) are subspaces of the same Euclidean space. Is this space Rn, Rk, or Rm?"

Dom(T2*T1) = Dom(T1) = R^n, so ker(T2*T1) and ker(T1) must both be contained in R^n.

"Prove that ker(T1) belongs to the subset of ker(T2 * T1)"

This is less trivial.

Let v be an element of ker(T1)

=> T1(v) = 0

=> T2*T1(v) = T2(0) = 0

=> ker(T2*T1) is contained in ker(T1)