Linear algebra problem related to vector subspace

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SUMMARY

The discussion focuses on solving a linear algebra problem involving the vector subspace X defined as X = {(x1, x2, x2 − x1, 3x2): x1, x2 ∈ R}. The basis for X consists of the vectors (1, 1, 0, 3) and (1, 2, 1, 6), leading to a dimension of 2. The kernel of the function f is identified as ker f = {0}, and the image of f is the entire subspace X. The function f is not a bijection, as it does not satisfy the criteria for one-to-one mapping.

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  • Understanding of vector spaces and subspaces
  • Familiarity with linear transformations and their properties
  • Knowledge of kernel and image of a linear transformation
  • Ability to determine linear independence of vectors
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  • Learn about linear transformations and how to compute their kernels and images
  • Explore the concept of linear independence and methods to prove it
  • Investigate bijections in the context of linear mappings
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Homework Statement


X ={(x1,x2,x2 −x1,3x2):x1,x2 ∈R}
f(x1,x2,x2 −x1,3x2)=(x1,x1,0,3x1)
1. Find a basis for X.
2. Find dim X.
3. Find ker f and I am f
4. Find bases for ker f and I am f
5. Is f a bijection? Why?
6. Find a diagonal matrix for f.

Homework Equations

The Attempt at a Solution


1. Put x1=x2=1: (1, 1, 0, 3) and Put x1=1 and x2=2: (1, 2, 1, 6)
2. Dim X = 2 as there are two vectors
3. Ker f = 0, I am f = X
4. (0,0,0,0)
5. I guess no, but do not know how to explain
6. No idea
 
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1. Looks like a good guess, but you will need to prove that the two vectors you found span X.

2. Here you will also need to prove that the two vectors you found in 1 are linearly independent.

3. ker f is a set, not a vector or a number. If you meant ker f={0}, where 0 is the zero vector, then the answer is wrong since e.g. f(0,1,1,3)=0. Your answer for I am f is wrong too, since (1,2,1,6) is in X but not in I am f.
 

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