Linear Algebra: Prove that the set of invertible matrices is a Subspace

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SUMMARY

The set U = {A | A ∈ ℝⁿ, A is invertible} is not a subspace of ℝⁿ, the space of all n x n matrices. This conclusion is reached by demonstrating that U is not closed under vector addition or scalar multiplication, as evidenced by the identity matrix being included in U while 0*I and I + (-I) are not. The discussion emphasizes that without defining alternative operations for addition and multiplication, the standard definitions suffice for the proof.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically subspaces.
  • Familiarity with matrix operations, including addition and scalar multiplication.
  • Knowledge of invertible matrices and their properties.
  • Basic comprehension of vector spaces in ℝⁿ.
NEXT STEPS
  • Study the properties of vector spaces and subspaces in linear algebra.
  • Learn about invertible matrices and their role in linear transformations.
  • Explore alternative definitions of vector operations in abstract algebra.
  • Investigate the implications of closure properties in vector spaces.
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Students of linear algebra, mathematics educators, and anyone involved in theoretical mathematics or matrix theory will benefit from this discussion.

Millacol88
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Homework Statement


Is U = {A| A \in nn, A is invertible} a subspace of nn, the space of all nxn matrices?

The Attempt at a Solution


This is easy to prove if you assume the regular operations of vector addition and scalar multiplication. Then the Identity matrix is in the set but 0*I and I + (-I) are not, so its not closed under vector addition or scalar multiplication. Is there a way to prove this without assuming the usual vector operations?
 
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Millacol88 said:

Homework Statement


Is U = {A| A \in nn, A is invertible} a subspace of nn, the space of all nxn matrices?

The Attempt at a Solution


This is easy to prove if you assume the regular operations of vector addition and scalar multiplication. Then the Identity matrix is in the set but 0*I and I + (-I) are not, so its not closed under vector addition or scalar multiplication. Is there a way to prove this without assuming the usual vector operations?

Not until they tell you what the alternative addition and multiplication operations are. I don't think you have to worry about that. Your examples are just fine.
 
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