Question regarding linear transformation

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.
  • #1
Mathman23
254
0
Hi

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

I'm suppose to determain that T maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]

I was told by my professor about the following theorem.

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

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

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:

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

A being the standard matrix of the linear transformation.

A can also be written:

[tex]\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 ][/tex]

This is equal to:


[tex]\begin{equation}\nonumber
x + 4z = 0
\end{equation}[/tex]

[tex]\begin{equation}\nonumber
2x + y + 6z = 0
\end{equation}[/tex]

x in equation 1 can be written as [tex]x = -4z[/tex]

If I insert that x into equation 2 then I get

[tex]-2z + y = 0[/tex]

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

Sincerley and many thanks in advance

Fred
 
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  • #2
Actually this: "I have a linear transformation T which maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]
a A is the standard matrix for the linear transformation.

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

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 [tex]\mathbb{R}^2[/tex],there exist x in [tex]\mathbb{R}^3[/tex] 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.
 
  • #3


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.

 

Related to Question regarding linear transformation

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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