Is a Set of Invertible Matrices a Subspace?

[Maxwhale]

Homework Statement

S is a subset of vector space V.

If V = 2x2 matrix and S ={A | A is invertible}

a) is S closed under addition?
b) is S closed under scalar multiplication?

Homework Equations

The Attempt at a Solution

For non singular 2x2 matrices, S is not closed under addition. but I am not quite sure about invertible 2x2 matrix.

Say, A = [1 0]
[0 1]

So, if we add A + A, it is still invertible, so it is closed under addition. But does my statement lose the generality?

 

[Office_Shredder]

Your statement does lose the generality of the argument. Similarly I could argue if A is

[1 0]
[0 0]

and B is

[0 1]
[0 0]

A+B is non-invertible, hence non-singular matrices are closed under addition. But you know that's false. To prove the set of invertible matrices is closed under addition, you need to prove that given ANY A and B (not just a single example) A + B is invertible. Alternatively, you can find a single counterexample to prove that it is not closed under + (since if there exists A,B such that A+B is not invertible, then it is not true that A+B is invertible for all A,B)

 

[Dick]

Homework Statement

S is a subset of vector space V.

If V = 2x2 matrix and S ={A | A is invertible}

a) is S closed under addition?
b) is S closed under scalar multiplication?

Homework Equations

The Attempt at a Solution

For non singular 2x2 matrices, S is not closed under addition. but I am not quite sure about invertible 2x2 matrix.

Say, A = [1 0]
[0 1]

So, if we add A + A, it is still invertible, so it is closed under addition. But does my statement lose the generality?

What's the difference between 'non-singular' and 'invertible'? Aren't they the same thing?