Help with a linear algebra problem

[kant]

1)

Any fuction T: V->W . V and W are vector spaces over the field of rational numbers. The fuction T is called additive if T(x+y)=T(x)+ T(y).

Proof that any function T from v to w are additive, then it must be a linear transformation.



2) Let T:V->W be linear

prove that if T is 1 to 1 IFF T carries linear indep subset in V to Linear, indep subset in W.

 

[JasonRox]

1)

Any fuction T: V->W . V and W are vector spaces over the field of rational numbers. The fuction T is called additive if T(x+y)=T(x)+ T(y).

Proof that any function T from v to w are additive, then it must be a linear transformation.



2) Let T:V->W be linear

prove that if T is 1 to 1 IFF T carries linear indep subset in V to Linear, indep subset in W.

1. Hint - Write the definition of a linear transformation right next to the question.

2. Hint - Assume it maps a lin. ind. subset in V to a lin. dep. subset to W.

That's only one direction, but now you have to go the other way too. (if and only if)

 

[tacman]

1) To prove that a function T from V to W is a linear transformation, we need to show that it satisfies two properties: additivity and scalar multiplication. We are given that T is additive, which means that T(x+y) = T(x) + T(y) for any vectors x and y in V. Now, let's consider scalar multiplication. Since V and W are vector spaces over the field of rational numbers, we can multiply vectors by rational numbers. Therefore, if we have a vector v in V and a rational number r, we can define the vector rv in V as rv = r*v, where * is the scalar multiplication operation. Similarly, we can define scalar multiplication in W. Now, if we multiply a vector v in V by a rational number r and apply the function T, we should get the same result as first applying T to v and then multiplying by r. In other words, T(rv) = rT(v). This is true because T is a linear transformation and satisfies the property of scalar multiplication. Therefore, we have shown that T is both additive and satisfies scalar multiplication, and thus, it is a linear transformation.

2) To prove that if T is 1 to 1, then it carries linearly independent subsets in V to linearly independent subsets in W, we can use the contrapositive statement. In other words, we can prove that if T carries linearly dependent subsets in V to linearly dependent subsets in W, then T is not 1 to 1. Let's assume that T carries a linearly dependent subset {v1, v2, ..., vn} in V to a linearly dependent subset {w1, w2, ..., wn} in W. This means that there exist scalars a1, a2, ..., an (not all zero) such that a1w1 + a2w2 + ... + anwn = 0. Now, since T is a linear transformation, we know that T(av1 + bv2 + ... + cvn) = aT(v1) + bT(v2) + ... + cT(vn) = a0 + b0 + ... + c0 = 0, where a, b, c are scalars and T(vi) = wi. This means that the set {av1 + bv2 + ... + cvn} is a linearly dependent subset in V. Since T carries a