# Question regarding linear transformation

• Mathman23
In summary, the conversation discusses a linear transformation T from \mathbb{R}^3 to \mathbb{R}^2 and the standard matrix A for this transformation. The goal is to prove that T maps \mathbb{R}^3 to \mathbb{R}^2. The professor introduces a theorem stating that T maps \mathbb{R}^n to \mathbb{R}^m if and only if A spans \mathbb{R}^m. The conversation then goes on to show how to apply this theorem to the given problem.
Mathman23
Hi

I have a linear transformation T which maps $$\mathbb{R}^3 \rightarrow \mathbb{R}^2$$ a A is the standard matrix for the linear transformation.

I'm suppose to determain that T maps $$\mathbb{R}^3 \rightarrow \mathbb{R}^2$$

I was told by my professor about the following theorem.

If $$T:\mathbb{R}^n \rightarrow \mathbb{R}^m$$ is a linear transformation and A is the standard matrix for T. Then

a/ T maps $$\mathbb{R}^n \rightarrow \mathbb{R}^m$$ if and only if A $$\mathrm{span} \{ \mathbb{R}^m \}$$

b/ T is one-to-one if and only if columns of A are linearly independent.

If I then apply (a) from to my problem:

$$A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ]$$

A being the standard matrix of the linear transformation.

A can also be written:

$$\left[ \begin{array}{cc} 1 \\2 \end{array} \right ] \mathrm{x} + \left[ \begin{array}{cc} 0 \\1 \end{array} \right ] \mathrm{y} + \left[ \begin{array}{cc} 4 \\6 \end{array} \right ] \mathrm{z} = \left[ \begin{array}{cc} 0 \\0 \end{array} \right ]$$

This is equal to:

$$\nonumber x + 4z = 0$$

$$\nonumber 2x + y + 6z = 0$$

x in equation 1 can be written as $$x = -4z$$

If I insert that x into equation 2 then I get

$$-2z + y = 0$$

But what do I then do to prove point (a) in the theorem ?

Sincerley and many thanks in advance

Fred

Actually this: "I have a linear transformation T which maps $$\mathbb{R}^3 \rightarrow \mathbb{R}^2$$
a A is the standard matrix for the linear transformation.

I'm suppose to determain that T maps $$\mathbb{R}^3 \rightarrow \mathbb{R}^2$$" doesn't quite make sense! I assume you mean simply that T is from$$\mathbb{R}^3$$to $$\mathbb{R}^2$$ and you want to prove this is an "onto" mapping.
Okay, you are given A= $$A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ]$$

I don't understand why you then set Ax= 0: that would prove that the columns are linearly independent which is your professor's (b), not (a) (and this is clearly not one to one). To show that A is "onto", you need to show that for all y in $$\mathbb{R}^2$$,there exist x in $$\mathbb{R}^3$$ so that Ax= y. The simplest way to do that is to row-reduce A. If the second row does not reduce to all 0s, it's true.

Hi Fred,

To prove point (a) in the theorem, you need to show that the span of the columns of A is equal to \mathbb{R}^m. In other words, you need to show that any vector in \mathbb{R}^m can be written as a linear combination of the columns of A.

In your case, since A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ], the span of the columns of A will be all possible linear combinations of the three columns. This means that any vector in \mathbb{R}^2 can be written as a linear combination of the columns of A. For example, the vector \left[ \begin{array}{cc} 2 \\ 3 \end{array} \right ] can be written as 2\left[ \begin{array}{cc} 1 \\ 2 \end{array} \right ] + 3\left[ \begin{array}{cc} 0 \\ 1 \end{array} \right ] + 0\left[ \begin{array}{cc} 4 \\ 6 \end{array} \right ]. This shows that the span of the columns of A is equal to \mathbb{R}^2, and therefore, T maps \mathbb{R}^3 \rightarrow \mathbb{R}^2.

I hope this helps. Let me know if you have any other questions.

## 1. What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space to another, preserving the basic structure of the original space. It is often represented as a matrix multiplication and can be used to describe many real-world processes, such as rotations, reflections, and scaling.

## 2. How do you determine if a transformation is linear?

A transformation is linear if it follows two basic rules:
1. The transformation of a sum is equal to the sum of the transformations: T(u + v) = T(u) + T(v)
2. The transformation of a scalar multiple is equal to the scalar multiple of the transformation: T(cu) = cT(u)
If these two rules are satisfied, the transformation is considered linear.

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

A linear transformation preserves the basic structure of the original vector space, while a nonlinear transformation does not. This means that a linear transformation will maintain parallel lines as parallel and will not change the origin, while a nonlinear transformation can distort the shape of objects and alter the origin.

## 4. How is the concept of linear transformation used in real-world applications?

Linear transformations are used in various fields of science, such as physics, engineering, economics, and computer graphics. They can be used to model and analyze real-world processes, such as the movement of objects, changes in economic systems, and transformations of images in computer graphics.

## 5. Can a linear transformation have a negative determinant?

Yes, a linear transformation can have a negative determinant. The determinant of a linear transformation is used to determine the scaling factor of the transformation. A negative determinant indicates that the transformation results in a reflection or a change in orientation, but it is still considered a linear transformation.

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