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

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Sanglee
#1
Aug4-11, 12:41 AM
P: 6
1. The problem statement, all variables and given/known data

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.

2. Relevant equations



3. 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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vela
#2
Aug4-11, 02:22 AM
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P: 11,769
Quote Quote by Sanglee View Post
1. The problem statement, all variables and given/known data

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.

2. Relevant equations



3. 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.
Sanglee
#3
Aug4-11, 03:53 AM
P: 6
So clear, easy to understand. Thanks!


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