Confirm Subspace: 2x2 Matrices B Where B[1 -1]^t=0

[pyroknife]

Which of the following is a subspace of M_2x2 (the vector space of 2x2 matrices. and explain why or why not:
1) Set of 2x2 matrices A such that det(A)=1
2) set of 2x2 matrices B such that B[1 -1]^t=0 vector
To check if something is a subspace I must satisfy 3 conditions (applied for matrix A):
1) 0 matrix is A
2) If U and V are in A then U+V is in A
3) if V is in A then cV is in A for some scalar c.
The above is analogous for matrix B.

For 1) Set of 2x2 matrices A such that det(A)=1
The 0 matrix is not in this set because the determinant is 0 which ≠1, thus the set of 2x2 matrices A is not a subspace.

Is this correct?
For 2) set of 2x2 matrices B such that B[1 -1]^t=0 vector
The 0 matrix is in this set because the matrix 2x2 consisting of all 0s multiplied by [1 -1]^t is =0.

Now I want to make sure I'm correctly applying the latter 2 conditions.
If U and V are in this set, then the following is true.
If U*[1 -1]^t=0 and V*[1 -1]^t=0
U+V=(U+V)[1 -1]^t. Since U and V are 0, U+V=0. thus U+V=(0+0)[1 -1]^t.=0, thus U+V is in B?

For condition 3: if V1*[1 -1]^t=0 is in the set, then cV must be in the set for it to be a subspace.
cV1*[1 -1]^t=c*0*[1 -1]^t=0, thus cV is in the set?

Thus the set of 2x2 matrices B such that B[1 -1]^t=0 vector is a subspace.

 

[haruspex]

If U and V are in B, then they're of the following form
If U=U1*[1 -1]^t=0 and V=V1*[1 -1]^t=0

First, B is not the set of interest. It is used as an example of one of the matrices. No letter has been assigned to the set, so let's call it S.
If U is in S it means U*[1 -1]^t=0. U is not of the form U1*[1 -1]^t, and it is not usually 0.

 

[pyroknife]

First, B is not the set of interest. It is used as an example of one of the matrices. No letter has been assigned to the set, so let's call it S.
If U is in S it means U*[1 -1]^t=0. U is not of the form U1*[1 -1]^t, and it is not usually 0.

Thanks.

I edited my original post. Does it look better?

 

[micromass]

Thanks.

I edited my original post. Does it look better?

Please do not edit your original post and just reply.

 

[haruspex]

If U and V are in this set, then the following is true.
If U*[1 -1]^t=0 and V*[1 -1]^t=0

Yes.

U+V=(U+V)[1 -1]^t.

No it doesn't

Since U and V are 0, U+V=0.

No they aren't

thus U+V=(0+0)[1 -1]^t.=0,

No it isn't.
Stop and think... what do you have to prove about U+V that would imply it is in S?

 

[pyroknife]

Yes.

No it doesn't

No they aren't

No it isn't.
Stop and think... what do you have to prove about U+V that would imply it is in S?

I have to prove that the addition of two things in S must be in S in order for it to be a subspace.

Instead of U+V=(U+V)[1 -1]^t. , I should've written:
U*[1 -1]^t +V*[1 -1]^t=0=(U+V)*[1 -1]^t=0
(0+0)*[1 -1]^t=0
Thus U*[1 -1]^t +V*[1 -1]^t is in S.

Is this the right idea?

 

[haruspex]

I have to prove that the addition of two things in S must be in S in order for it to be a subspace.

Instead of U+V=(U+V)[1 -1]^t. , I should've written:
U*[1 -1]^t +V*[1 -1]^t=0=(U+V)*[1 -1]^t=0

Yes, and if you just rearrange that sequence you will have a logical proof. But what has the next line to do with anything?

(0+0)*[1 -1]^t=0
Thus U*[1 -1]^t +V*[1 -1]^t is in S.

No, you're not trying to prove U*[1 -1]^t +V*[1 -1]^t is in S. You want to show U+V is in S.