Linear algebra - Image and Kernel

In summary: Similarly, the second and third columns are the results of applying T to (0,1,0)T and (0,0,1)T, respectively.In summary, the question provides a linear transformation T on a 3-dimensional vector space over F, and asks for the matrix of T, as well as the basis of the kernel and image of T. The solution involves ordering the basis vectors in columns, row-reducing the resulting matrix to find the span of the image, and solving a matrix equation to find the kernel.
  • #1
sincera4565
1
0

Homework Statement



Let V be a 3 dim vector space over F and e_1 e_2 and e_3 be those fix basis
The question provide us with the linear transformation T[itex]\in[/itex] L(V) such that
T(e_1) = e_1 + e_2 - e_3
T(e_2) = e_2 - 3e_3
T(e_3) = -e_1 -3e_2 -2e_3

we are ask to find the matrix of T and the basis of ker(T) and Im(T)

2. The attempt at a solution

I think I find the matrix right
where the matrix of T should be
1 0 -1
1 1 -3
-1 -3 -2

but the problem is I am not sure how can i find the ker(T) and Im(T)
 
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  • #2
Few pointers:
o The matrix you wrote down is wrong. Look for the proper way to order the coordinates of the basis vectors.
o For Im(T): You have to find the span of vectors
o For Ker(T): You need to solve a matrix equation
 
  • #3
MednataMiza said:
Few pointers:
o The matrix you wrote down is wrong. Look for the proper way to order the coordinates of the basis vectors.
i'm not seeing this. it looks correct to me.
o For Im(T): You have to find the span of vectors
o For Ker(T): You need to solve a matrix equation
if one uses row-reduction, one could accomplish both at the same time.
 
  • #4
Am I wrong or one should order the vectors in columns not in rows ?
Finding the span is just row-reducing the matrix ...
 
  • #5
MednataMiza said:
Am I wrong or one should order the vectors in columns not in rows ?
Finding the span is just row-reducing the matrix ...

it appears that is what has been done.

T(e1) = e1 + e2 - e3,

that is: T((1,0,0)T) = 1(1,0,0)T + 1(0,1,0)T + (-1)(0,0,1)T

= (1,1,-1)T, which appears to be the first column of sincera4565's matrix.
 

Related to Linear algebra - Image and Kernel

1. What is the difference between an image and a kernel in linear algebra?

The image of a linear transformation is the set of all possible outputs that can be produced by applying the transformation to a given set of inputs. It is also known as the range of the transformation. On the other hand, the kernel of a linear transformation is the set of all inputs that produce a zero output when the transformation is applied to them. It is also known as the null space of the transformation.

2. How are the image and kernel related in linear algebra?

The image and kernel of a linear transformation are related by the rank-nullity theorem, which states that the sum of the dimensions of the image and kernel is equal to the dimension of the domain of the transformation. In other words, the dimensions of the image and kernel add up to the total number of inputs that the transformation can be applied to.

3. Can the image and kernel of a linear transformation be empty?

Yes, it is possible for the image and kernel of a linear transformation to be empty. This occurs when the transformation is not onto (does not map to all possible outputs) or not one-to-one (maps different inputs to the same output). In these cases, the image and/or kernel will have a dimension of 0, meaning that they do not contain any vectors.

4. How can the image and kernel of a linear transformation be used in applications?

The image and kernel of a linear transformation are useful in many applications, such as image and signal processing, data compression, and machine learning. They can be used to analyze and manipulate data, reduce redundancy, and identify important features or patterns in a dataset.

5. Can the image and kernel of a linear transformation be changed by scaling or rotating the input?

No, scaling or rotating the input will not change the image or kernel of a linear transformation. This is because linear transformations preserve the structure of the vector space, so the image and kernel will remain the same regardless of how the inputs are scaled or rotated.

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