(LinearAlgebra) all 2x2 invertible matrices closed under addition?

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SUMMARY

The set of all 2x2 invertible matrices is not closed under addition. A definitive counter-example is provided: the sum of the identity matrix I and its additive inverse -I results in the zero matrix, which is not invertible. Therefore, while both I and -I are elements of the vector space V of invertible matrices, their sum does not belong to V, confirming that V is not closed under addition.

PREREQUISITES
  • Understanding of vector spaces and their properties
  • Knowledge of matrix operations, specifically addition
  • Familiarity with the concept of invertible matrices
  • Basic linear algebra concepts, including the identity matrix
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  • Learn about the implications of closure properties in vector spaces
  • Explore the characteristics of invertible matrices and their operations
  • Investigate additional examples of non-closed sets in linear algebra
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Students and educators in linear algebra, particularly those studying vector spaces and matrix theory, will benefit from this discussion.

Sanglee
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Homework Statement



Suppose V is a vector space.
Is the set of all 2x2 invertible matrices closed under addition? If so, please prove it. If not, please
provide a counter-example.

Homework Equations





The Attempt at a Solution



well i know that what does it mean to be closed under addition. When V is closed under addition, if I suppose vector u and w are in the V, their addition u+w is also in the V, right?

The answer for the question is No.
A counter-example my professor provided is I+(-I)=0
I and (-I) are invertible, but their addition 0 is not invertible. and I know why it's not invertible.
But I don't figure out why it is not closed under addition,,.
If the addition is not invertible, does it mean that the addition is not in the V?
 
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Sanglee said:

Homework Statement



Suppose V is a vector space.
Is the set of all 2x2 invertible matrices closed under addition? If so, please prove it. If not, please
provide a counter-example.

Homework Equations





The Attempt at a Solution



well i know that what does it mean to be closed under addition. When V is closed under addition, if I suppose vector u and w are in the V, their addition u+w is also in the V, right?
Yes.
The answer for the question is No.
A counter-example my professor provided is I+(-I)=0
I and (-I) are invertible, but their addition 0 is not invertible. and I know why it's not invertible.
But I don't figure out why it is not closed under addition,,.
If the addition is not invertible, does it mean that the addition is not in the V?
Yes. V consists of only invertible matrices, so 0 is not an element in V. So you have u=I and w=-I are both in V, but their sum u+w=0 is not in V. Therefore V is not closed under addition.
 
So clear, easy to understand. Thanks!
 

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