How to Determine a Vector Not in the Range of a Linear Transformation?

  • Thread starter georgeh
  • Start date
In summary, the range of the linear operator defined by the equations W_1=x_1 - 2*x_2 + x_3, W_2=5*x_1-x_2+3 *x_3, W_3=4*x_1+x_2+2*x_3 is not all of R^3. This can be shown by using the linear transformation matrix T=[1,-2,1;5,-1,3;4,1,2] and reducing the augmented matrix T|W. The fact that there is a vector (10, 5, 6) that is not in the range of the transformation implies that the three equations are linearly dependent. This can be
  • #1
georgeh
68
0
The question asls:Show that the range of the linear operator defined by the equations:
W_1=x_1 - 2*x_2 + x_3
W_2=5*x_1-x_2+3 *x_3
W_3=4*x_1+x_2+2*x_3
is not all of R^3, and find a vector that is not in the range.
Well, we know T
T=[1,-2,1;5,-1,3;4,1,2]
I augment W
and we get
T|W=[1,-2,1,W_1;5,-1,3,W_2;4,1,2,W_3]
I do Reduce Echelon
I get zeros on the bottom
so i get
w_1-W_2+W_3=0
k, well, i chose a vector point, (10,5,6)
i get
10 != -1
not sure if this is correct way of solving it.
 
Physics news on Phys.org
  • #2
If I read your problem correctly, here's a hint. If there is a vector that is not in the range of the transformation, then what does that tell you about the linear dependence or independence of the three equations?
 
  • #3
we haven't covered linear independence yet.
 
  • #4
georgeh said:
The question asls:Show that the range of the linear operator defined by the equations:
W_1=x_1 - 2*x_2 + x_3
W_2=5*x_1-x_2+3 *x_3
W_3=4*x_1+x_2+2*x_3
is not all of R^3, and find a vector that is not in the range.
Well, we know T
T=[1,-2,1;5,-1,3;4,1,2]
I augment W
and we get
T|W=[1,-2,1,W_1;5,-1,3,W_2;4,1,2,W_3]
I do Reduce Echelon
I get zeros on the bottom
so i get
w_1-W_2+W_3=0
k, well, i chose a vector point, (10,5,6)
i get
10 != -1
not sure if this is correct way of solving it.

WHY did you choose (10, 5, 6) and HOW did you "get 10 != -1"?
 

What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations, linear functions, and their representations in vector spaces and matrices.

Why is linear algebra important?

Linear algebra is important because it provides a foundation for many other areas of mathematics, such as calculus and differential equations, and is used extensively in various fields including physics, engineering, computer science, and economics.

What are the basic concepts of linear algebra?

Some basic concepts of linear algebra include vectors, matrices, systems of linear equations, vector spaces, and linear transformations.

How do I solve a linear algebra problem?

To solve a linear algebra problem, you need to understand the basic concepts and principles of the subject and then apply appropriate methods and techniques, such as Gaussian elimination, matrix operations, and vector operations.

What are some real-world applications of linear algebra?

Linear algebra has many real-world applications, including image and signal processing, data compression, machine learning, cryptography, and computer graphics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
2K
  • Calculus and Beyond Homework Help
Replies
0
Views
449
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
617
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
830
  • Precalculus Mathematics Homework Help
Replies
32
Views
834
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top