? asked in Science & MathematicsMathematics · 1 decade ago

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!

2 Answers

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  • 1 decade ago
    Favorite Answer

    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.

    Just add equations:

            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.

  • 1 decade ago

    "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)

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