Is singular matrix is a subspace of vector space V?

[Maxwhale]

Homework Statement

S is a subset of vector space V,

If V is an 2x2 matrix and S={A|A is singular},

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

Homework Equations

S is a subspace of V if it is closed under addition and scalar multiplication.

The Attempt at a Solution

I tried to use the definition of sinularity. i.e. a matrix in not invertible. But could not decide if it was closed under addition and scalar multiplication.

 

[morphism]

Well, play around with examples of singular matrices. It's not too hard to find two that add up to an invertible one; in fact, you can find two singular matrices that add up to the identity.

 

[Maxwhale]

yeah, i can get an identity if i add [1 0; 0 0] and [0 0; 01]. So what does that insinuate? If we add two we get an invertible matrix, which implies that the solution to AX=B is unique. But how do I get to point where i can decide if they are closed under addition and scalar multiplication? Further help will be highly appreciated.

thank you

 

[morphism]

What does it mean for S to be closed under addition and scalar multiplication?

 

[Maxwhale]

if vec u and vec v are in S, vec u + vec v also should be in S (closed under addition)
for any scalar r, r(vec u) = r*vec(u) and lies in S(closed under scalar multiplication)

 

[morphism]

OK, now think about what you did: you added two things in S and got the identity matrix. What does this tell you about S?

 

[Maxwhale]

the identity matrix belongs to S. So u and v are closed under addition. Right?

 

[morphism]

Really -- does the identity matrix belong to S?

 

[Maxwhale]

ohh... so S is a singular, meaning that it is not invertible. The identity matrix is invertible, hence does not belong to S. I hope i got it right this time

 

[morphism]

Yup. So S isn't closed under addition.

Now, let's turn to scalar multiplication. Suppose we have a singular matrix A and we multiply it by a scalar r. Can rA ever be nonsingular (i.e. invertible)? Suppose it can be - what does this tell you about A?

 

[Maxwhale]

again, if we turn it into an invertible matrix, it would mean that A does not belong to S.
But i did not get an invertible matrix by multiplying a singular matrix with a scalar. Say A = [1 0; 0 0], if we multiply by r, we get, rA= [r 0; 0 0], which is still singular, right?

 

[HallsofIvy]

More succinctly, if det(A)= 0, then det(xA), for any scalar x, is what?

 

[FourierX]

we have not done determinant so far. But i believe in the above case, det(xA) should equal 0. I think so because det(xA) = x* det(A).

 

[HallsofIvy]

Yes, that's correct.