# Linear Algebra: basic proof about a linear transformation

• brushman
In summary, given a transformation T from V to W, where the dimension of V is n and the dimension of W is m, we can prove that if T is one-to-one and a set of vectors {v1, ..., vk} in V is linearly independent, then the set {T(v1), ..., T(vk)} in W is also linearly independent. This is because T being one-to-one means that the null space of T is 0, and since the vectors in V are linearly independent, the only solution to c1v1 + ... + ckvk = 0 is when all the coefficients are 0, which then translates to the set {T(v1), ..., T(vk)} being linear

## Homework Statement

Given:
T is a transformation from V -> W and the dim(V) = n and dim(W) = m (I think the dimensions were given for the purposes of another problem)

Prove:
If T is 1 to 1 and {v1, ..., vk} is a subspace of V that is L.I., then {T(v1), ..., T(vk)} is L.I in W.

## The Attempt at a Solution

First we form the set of transformations

A = {T(v1), ..., T(vk)}.

Since v1, ..., vk are L.I., none of these vectors are equal to each other.

Then since T is 1 to 1, T(vi) =/= T(vj) where vi and vj represent any vectors from {v1, ..., vk}.

Finally, this means that c1T(v1) + ... + ckT(vk) = 0 has only the trivial solution c1...ck = 0.

Therefore, {T(v1), ..., T(vk)} is L.I in W.

brushman said:

## Homework Statement

Given:
T is a transformation from V -> W and the dim(V) = n and dim(W) = m (I think the dimensions were given for the purposes of another problem)

Prove:
If T is 1 to 1 and {v1, ..., vk} is a subspace of V that is L.I., then {T(v1), ..., T(vk)} is L.I in W.
{v1, …, vk} isn't a subspace, and it doesn't make sense to say a subspace is linearly independent. {v1, …, vk} is simply a set of k linearly independent vectors in V.

## The Attempt at a Solution

First we form the set of transformations

A = {T(v1), ..., T(vk)}.

Since v1, ..., vk are L.I., none of these vectors are equal to each other.

Then since T is 1 to 1, T(vi) =/= T(vj) where vi and vj represent any vectors from {v1, ..., vk}.

Finally, this means that c1T(v1) + ... + ckT(vk) = 0 has only the trivial solution c1...ck = 0.
Just because two vectors aren't equal to each other doesn't mean they're linearly independent. The vectors (1,0,0) and (2,0,0) for example aren't equal to each other, but they are dependent.
Therefore, {T(v1), ..., T(vk)} is L.I in W.
You want to use the fact that T is linear. (You didn't say it was, but I assume that it's a given in the problem.) What can you say about the null space of T since it's one-to-one?

Thanks vela. Sorry about mis-wording the question, and yes, T is linear. Here is the revised version of my proof:Let A = {T(v1), ..., T(vk)} .

For {T(v1), ..., T(vk)} to be L.I., we must have the trivial solution for c1T(v1) + ... + ckT(vk) = 0 . Or, by linearity, T(c1v1 + ... + ckvk) = 0 .

Since T is 1 to 1, the null space of T must be 0. That is, T(0) = 0. This means that c1v1 + ... + ckvk = 0.

Because v1, ..., vk are L.I., all the c's must equal 0 (aka we have only the trivial solution).

Therefore we can conclude that {T(v1), ..., T(vk)} is L.I in W.

Perfect!

## 1. What is a linear transformation?

A linear transformation is a mathematical function that maps vectors from one vector space to another while preserving the operations of vector addition and scalar multiplication. In simpler terms, it is a way of transforming or manipulating vectors in a linear manner.

## 2. How do you prove that a function is a linear transformation?

To prove that a function is a linear transformation, you must show that it satisfies two properties: additive and homogeneous. Additive property means that the function of the sum of two vectors is equal to the sum of the function of each vector. Homogeneous property means that the function of a scalar multiple of a vector is equal to the scalar multiple of the function of the vector. If these two properties are satisfied, then the function is a linear transformation.

## 3. What is the difference between a linear transformation and a linear equation?

A linear transformation is a function that maps vectors from one vector space to another, while a linear equation is an algebraic equation with variables that have a degree of one. Linear transformations deal with vector spaces and operations on vectors, while linear equations deal with algebraic equations and solving for variables.

## 4. Can a linear transformation have more than one input and output?

Yes, a linear transformation can have multiple inputs and outputs. This is known as a multivariate linear transformation. The number of inputs is referred to as the dimension of the domain, and the number of outputs is referred to as the dimension of the co-domain.

## 5. How is linear algebra used in real life applications?

Linear algebra is used in various fields such as engineering, physics, economics, and computer science. It is used to solve systems of linear equations, analyze data and patterns, and transform data in image and signal processing. It also has applications in machine learning and artificial intelligence, helping to make predictions and decisions based on large datasets.