Math Problem Statement
what is the dimension and base
Solution
The problem involves understanding subspaces of the vector space , consisting of all matrices. Let's analyze the subspaces and and determine their dimensions and bases.
Space
The subspace is defined as:
This condition adds one linear constraint to the six entries of a general matrix. A linear constraint reduces the dimension of the space by 1. Since (6 entries in a matrix), the dimension of is:
Basis of
To construct a basis, start with the general matrix:
b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{pmatrix},$$ and impose the constraint $$b_{11} + b_{12} + b_{13} = b_{21} + b_{22} + b_{23}$$. Using the constraint, one entry can be expressed in terms of others (e.g., $$b_{11} = b_{21} + b_{22} + b_{23} - b_{12} - b_{13}$$). The remaining 5 entries are free, forming a basis of 5 independent matrices. --- ### Space $$W$$ The subspace $$W$$ consists of all matrices $$A = (a_{ij}) \in V$$ such that: 1. The sum of all entries of $$A$$ is zero: $$\sum a_{ij} = 0$$. 2. A specific linear relation is satisfied: $$\begin{pmatrix} a_{13} \\ a_{23} \end{pmatrix} = 4 \begin{pmatrix} a_{12} \\ a_{22} \end{pmatrix}.$$ #### Analyzing Constraints - The first condition adds one linear constraint. - The second condition relates $$a_{13}$$ and $$a_{23}$$ to $$a_{12}$$ and $$a_{22}$$, introducing two more linear constraints. This means $$W$$ is defined by $$3$$ independent constraints. Starting with $$\dim(V) = 6$$, the dimension of $$W$$ is: $$\dim(W) = 6 - 3 = 3.$$ #### Basis of $$W$$ To construct a basis for $$W$$, use the constraints to express 3 entries in terms of the remaining 3 free entries. For example: - Use $$\sum a_{ij} = 0$$ to express one entry in terms of others. - Use the linear relations for $$a_{13}$$ and $$a_{23}$$ to eliminate two more variables. The resulting 3 free variables correspond to independent matrices forming the basis. --- ### Intersection $$U \cap W$$ If needed, further analysis can find the intersection of $$U$$ and $$W$$, which involves solving all constraints simultaneously. --- ### Final Answer - **Dimension of $$U$$:** $$5$$. - **Dimension of $$W$$:** $$3$$. - **Bases:** Constructed explicitly by solving the constraints. Would you like a step-by-step construction of the bases? Let me know! --- ### Related Questions 1. What is the geometric interpretation of the constraint on $$U$$? 2. How do we determine the basis of $$U$$ explicitly? 3. Can we verify the constraints for $$W$$ geometrically? 4. What is the dimension of $$U \cap W$$, and how is it calculated? 5. How does changing the constraints on $$W$$ affect its dimension? ### Tip Always check the rank of the constraints when determining dimensions of subspaces, as they reveal the number of independent restrictions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Subspaces of Vector Spaces
Matrix Spaces
Linear Constraints
Formulas
Dimension formula: dim(U) = dim(V) - number of independent constraints
Matrix sum constraint: b_{11} + b_{12} + b_{13} = b_{21} + b_{22} + b_{23}
Linear relations in entries: a_{13} and a_{23} expressed in terms of a_{12} and a_{22}
Theorems
Rank-Nullity Theorem
Linear Independence and Basis Theorem
Suitable Grade Level
University-level Linear Algebra