Linear Algebra - Transformation / operator

In summary: Thanks again!In summary, to find a linear operator T:C^3→C^3 over C such that T is a projection with R(T)=span{(1,1,1)} and N(T)=span{(1,1,0);(0,1,1)}, all three rows of the T matrix must be identical and can be multiplied by a constant.
  • #1
Samuel Williams
20
3

Homework Statement


Let T:V→V be a linear operator on a vector space V over C:

(a) Give an example of an operator T:C^2→C^2 such that R(T)∩N(T)={0} but T is not a projection

(b) Find a formula for a linear operator T:C^3→C^3 over C such that T is a projection with R(T)=span{(1,1,1)} and N(T)=span{(1,1,0);(0,1,1)}

Homework Equations



If T is a projection, then T^2=T
R(T) is the range of T (I think it is also the image), N(T) is the null space or kernel of T.

The Attempt at a Solution



I don't know how to create one where R(T)∩N(T)={0}. I sort of figured out by using [2 0 , 0 1/2]
As for (b), I try setting T(1,1,0)=(0,0,0) and T(0,1,1)=(0,0,0) but I'm not sure if this is correct or what to do afterwards.
 
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  • #2
For part (a), the R(T) ∩ N(T) = {0} implies that only the zero vector maps to zero. So, your idea to use
##\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}## was valid. The identity matrix satisfies R(T) ∩ N(T) = {0}, but it is a projection, so using non-ones on the diagonal does the trick.
For part (b) you are right to set the conditions for the null space. The fact that the range is the span of (1,1,1) means that the output in all three columns should be the same, i.e. T( x, y, z) = (f(x,y,z), f(x,y,z), f(x,y,z)).
Unless I am misinterpreting what you wrote, and you meant that R(T) = span{ (1,0,0), (0,1,0), (0,0,1) } which would be the entire space.
 
  • #3
These are finite vector spqces, just represent the transformationa with 2x2 / 3x3 matrices and see what constraints you obtain from the requirements.
 
  • #4
RUber said:
For part (a), the R(T) ∩ N(T) = {0} implies that only the zero vector maps to zero. So, your idea to use
##\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}## was valid. The identity matrix satisfies R(T) ∩ N(T) = {0}, but it is a projection, so using non-ones on the diagonal does the trick.
For part (b) you are right to set the conditions for the null space. The fact that the range is the span of (1,1,1) means that the output in all three columns should be the same, i.e. T( x, y, z) = (f(x,y,z), f(x,y,z), f(x,y,z)).
Unless I am misinterpreting what you wrote, and you meant that R(T) = span{ (1,0,0), (0,1,0), (0,0,1) } which would be the entire space.

You are correct in saying that the range is the span of (1,1,1), but I am not sure what to do after this.
 
  • #5
As Orodruin mentioned, your goal is to make a 3x3 matrix for the transformation.
##\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3}\\a_{2,1} & a_{2,2}& a_{2,3}\\a_{3,1} & a_{3,2}& a_{3,3} \end{pmatrix}
\begin{pmatrix} x \\ y \\z\end{pmatrix} = \begin{pmatrix} b_1\\b_2\\b_3\end{pmatrix}##
Your conditions are that b_1 = b_2 = b_3 to be in the span of (1,1,1). What does this mean about your three rows of the T matrix?
Add that to your null space requirements, you should be able to provide a general form in terms of some constant value.
 
  • #6
RUber said:
As Orodruin mentioned, your goal is to make a 3x3 matrix for the transformation.
##\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3}\\a_{2,1} & a_{2,2}& a_{2,3}\\a_{3,1} & a_{3,2}& a_{3,3} \end{pmatrix}
\begin{pmatrix} x \\ y \\z\end{pmatrix} = \begin{pmatrix} b_1\\b_2\\b_3\end{pmatrix}##
Your conditions are that b_1 = b_2 = b_3 to be in the span of (1,1,1). What does this mean about your three rows of the T matrix?
Add that to your null space requirements, you should be able to provide a general form in terms of some constant value.

Would I then be correct in saying that all three rows must be identical? If so, using the null space constraints should then give a matrix like :
\begin{pmatrix} 1 & -1 & 1 \\ 1 & -1 & 1 \\ 1 & -1 & 1 \end{pmatrix}
If I am not mistaken?
 
  • #7
That's right. And notice that you can multiply that matrix by any constant and it will still satisfy the requirements in your problem.
 
  • #8
Thanks to you both! I appreciate your input and I also appreciate that you helped me through rather than just answering the question. I definitley gained a better understanding
 
  • Like
Likes RUber

1. What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space to another, while preserving the basic structure of the original space. This means that the transformation must follow certain rules, such as preserving addition and scalar multiplication.

2. What is a linear operator?

A linear operator is a special type of linear transformation that maps a vector space to itself. It is often represented as a matrix and can be thought of as a function that takes in a vector and outputs another vector.

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

The main difference between a linear transformation and a linear operator is the type of vector spaces they map to. A linear transformation can map to any vector space, while a linear operator maps to the same vector space. Another way to think about it is that a linear operator is a specific type of linear transformation.

4. How do you determine if a transformation is linear?

To determine if a transformation is linear, you can check if it follows the two main rules of linearity: preserving addition and scalar multiplication. This means that the transformation must satisfy the equations f(u+v) = f(u) + f(v) and f(ku) = kf(u), where u and v are vectors and k is a scalar.

5. What are some real-life applications of linear transformations and operators?

Linear transformations and operators have many applications in fields such as physics, engineering, and computer science. For example, they can be used to model physical systems, analyze data, and develop algorithms for image and signal processing. They are also essential in understanding and solving systems of linear equations, which have many real-world applications.

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