Math 250b Quiz 6 Due 6/22/15 Name
Math 250b Quiz 6due 62215name
Determine a basis for the row space, a basis for the column space and a basis for the null space of the given matrix A . Find the dimension of each space. ຠຠຠ໠๠ઠઠઠૠé - - - - = A .
Let { } ) 1 , 1 , 1 ( ), 5 , 2 , 1 ( ), 5 , 3 , 2 ( ), 1 , 1 , 1 ( - - - = S .
- Does the set S span R 3 ?
- Express the vector ) 11 , 4 , 1 ( - = v r , if possible, as a linear combination of the vectors from S .
- Determine whether the set S is linearly dependent or linearly independent in R 3 .
- Is S a basis for R 3 ? Justify your answer.
Paper For Above instruction
In this paper, we will thoroughly analyze the matrix A to determine bases for its row space, column space, and null space, along with their respective dimensions. Additionally, we will explore the properties of the set S, including whether it spans R³, whether a specific vector can be expressed as a linear combination of vectors from S, the linear dependence or independence of S, and whether S forms a basis for R³.
Analysis of Matrix A: Bases and Dimensions
Given the matrix A (though not explicitly provided in the prompt, the typical process involves row reducing A to echelon form). To find a basis for the row space of A, one must identify the set of non-zero rows in the row echelon form of A. These non-zero rows will be linearly independent and span the row space. The dimension of the row space corresponds to the number of such non-zero rows.
For the column space, the basis consists of the original columns of A that correspond to the leading entries in the row echelon form. The number of such columns gives the dimension of the column space, which is known as the rank of A.
The null space basis involves solving the homogeneous system Ax = 0 to find the free variables and thus express the null space basis vectors. The dimension of the null space, or nullity, is given by the total number of variables minus the rank of A as per the rank-nullity theorem.
Properties of Set S in R³
The set S, consisting of vectors s₁ = (1, 1, 1), s₂ = (5, 2, 1), s₃ = (5, 3, 2), and s₄ = (1, 1, 1), needs to be examined for spanning R³. For S to span R³, its vectors must be linearly independent and cover the entire space, meaning any vector in R³ can be expressed as their linear combination.
To check whether the vector v = (11, 4, 1) can be expressed as a linear combination of vectors in S, one sets up the equation:
αs₁ + βs₂ + γs₃ + δs₄ = v,
and attempts to solve for the scalars α, β, γ, δ. The solvability indicates whether v can be represented as a combination of vectors in S.
The linear dependence or independence of S is determined by examining the rank of matrix formed by the vectors as columns or rows. If the rank is less than the number of vectors, S is linearly dependent; otherwise, it is independent.
Finally, S is a basis for R³ if it is linearly independent and spans R³.
Conclusion
Without explicit matrix entries, the general process involves row reduction and solving linear systems to determine the bases and dimensions. The properties of S rest on linear algebra principles: span, linear dependence, and basis criteria. Accurate calculations for the specific matrix A and set S are necessary to provide definitive answers.
References
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