What is the Cardinality and Dimension of \mathbb{Z}^{3}_{7}?

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SUMMARY

The cardinality of the vector space \(\mathbb{Z}^{3}_{7}\) over \(\mathbb{Z}_{7}\) is 343, derived from the fact that each of the three components can take on 7 values (from 0 to 6). The dimension of \(\mathbb{Z}^{3}_{7}\) is confirmed to be 3, with the basis defined as \(B = \{ (1,0,0), (0,1,0), (0,0,1) \}\). This establishes that \(\mathbb{Z}^{3}_{7}\) is a three-dimensional vector space over the field \(\mathbb{Z}_{7}\).

PREREQUISITES
  • Understanding of vector spaces
  • Familiarity with finite fields, specifically \(\mathbb{Z}_{7}\)
  • Knowledge of basis and dimension concepts in linear algebra
  • Basic operations in modular arithmetic
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  • Study the properties of finite fields, focusing on \(\mathbb{Z}_{p}\) for prime \(p\)
  • Explore vector space theory, particularly the concepts of basis and dimension
  • Learn about the applications of vector spaces in coding theory
  • Investigate the implications of cardinality in different vector spaces
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Students of linear algebra, mathematicians exploring finite fields, and educators teaching vector space concepts will benefit from this discussion.

jdstokes
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Find the cardinality and dimension of the vector space \mathbb{Z}^{3}_{7} over \mathbb{Z}_{7}.

\mathbb{Z}^{3}_{7} = \{ (a,b,c) \; | \; a,b,c \in \mathbb{Z}_{7} \}.

Then since \mathbb{Z}_{7} is a field 1 \cdot a = a \; \forall \; a, so B = \{ (1,0,0), (0,1,0) , (0,0,1) \} is a basis of \mathbb{Z}^{3}_{7}, so \dim \mathbb{Z}^{3}_{7} = 3. ans = 9, what the?

Thanks

James
 
Last edited:
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Hi again,

Does anyone have any clues on this one? I'm really stuck.

Thanks

James.
 
Are you sure you copied the problem correctly? Because 3 certainly seems to be the right answer.
 

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