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Let Vbe a vector space over (ℝ) of dimension 7 and T :V → ℝ be a non-zero linear transformation. Let W be a linear subsapce of V such that V = ker (T) ⊕ W, Where ker (T) denotes the null space of T. Then dimension of W is (Answer should be integer)___________.
Correct answer is '1'. Can you explain this answer?
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Let Vbe a vector space over () of dimension 7 and T :V →be a non-...
Understanding the Problem
To find the dimension of the subspace W given the conditions of the vector space V and the linear transformation T, we need to analyze the components involved.
Key Concepts
- Vector Space (V): It has a dimension of 7.
- Linear Transformation (T): It is non-zero, meaning it does not map all vectors to zero.
- Kernel of T (ker(T)): This is the null space of T, the set of all vectors that T maps to zero.
- Direct Sum (V = ker(T) ⊕ W): This means every vector in V can be uniquely expressed as a sum of a vector from ker(T) and a vector from W.
Applying the Rank-Nullity Theorem
The Rank-Nullity Theorem states that:
- Dimension of V = Dimension of ker(T) + Dimension of im(T)
Here, im(T), the image of T, is non-empty since T is non-zero.
Calculating Dimensions
- Let the dimension of ker(T) be k.
- Then, the dimension of im(T) will be 7 - k.
Since T is non-zero, im(T) must have a dimension of at least 1. Thus:
- 7 - k ≥ 1
- This implies k ≤ 6.
Finding Dimension of W
Given the direct sum:
- Dimension of V = Dimension of ker(T) + Dimension of W
- 7 = k + Dimension of W
Substituting the maximum value of k (which is 6):
- 7 = 6 + Dimension of W
- Dimension of W = 1.
Conclusion
The dimension of the subspace W is indeed 1, confirming the final answer is:
1
To find the dimension of the subspace W given the conditions of the vector space V and the linear transformation T, we need to analyze the components involved.
Key Concepts
- Vector Space (V): It has a dimension of 7.
- Linear Transformation (T): It is non-zero, meaning it does not map all vectors to zero.
- Kernel of T (ker(T)): This is the null space of T, the set of all vectors that T maps to zero.
- Direct Sum (V = ker(T) ⊕ W): This means every vector in V can be uniquely expressed as a sum of a vector from ker(T) and a vector from W.
Applying the Rank-Nullity Theorem
The Rank-Nullity Theorem states that:
- Dimension of V = Dimension of ker(T) + Dimension of im(T)
Here, im(T), the image of T, is non-empty since T is non-zero.
Calculating Dimensions
- Let the dimension of ker(T) be k.
- Then, the dimension of im(T) will be 7 - k.
Since T is non-zero, im(T) must have a dimension of at least 1. Thus:
- 7 - k ≥ 1
- This implies k ≤ 6.
Finding Dimension of W
Given the direct sum:
- Dimension of V = Dimension of ker(T) + Dimension of W
- 7 = k + Dimension of W
Substituting the maximum value of k (which is 6):
- 7 = 6 + Dimension of W
- Dimension of W = 1.
Conclusion
The dimension of the subspace W is indeed 1, confirming the final answer is:
1
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Community Answer
Let Vbe a vector space over () of dimension 7 and T :V →be a non-...
Let T:V → ℝ be a non zero linear transformation, then rank (T) = 1. Hence Ker(T) = 7 - 1=6
Given V = Ker(T)⊕W
⇒ dim(V) = dim(ker(T)) + dim(W)
⇒ 7 = 6+ dim(W)
⇒ dim(W) =1
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Let Vbe a vector space over () of dimension 7 and T :V →be a non-zero linear transformation. Let W be a linear subsapce of V such that V = ker (T)⊕ W, Where ker (T) denotes the null space of T. Then dimension of W is (Answer should be integer)___________.Correct answer is '1'. Can you explain this answer?
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