I just finished Exam 1 and I was checking the answers.
Exercise 4 item (c) asks what subspace S is spanned by all possible row reduced echelon forms matrices. I answered none, since they do not form a subspace (at least in my conception).
Proof (?) = The identity matrix is in the row reduced echelon form, so it would be in the subspace. If I multiply it by 3 I would get the matrix [3 0 0 ; 0 3 0 ; 0 0 3] wich is clearly not in the row reduced echelon form and consequently not in the S "subspace".
Am I missing something?

Question

I just finished Exam 1 and I was checking the answers. 
Exercise 4 item (c) asks what subspace S is spanned by all possible row reduced echelon forms matrices. I answered none, since they do not form a subspace (at least in my conception). 
Proof (?) = The identity matrix is in the row reduced echelon form, so it would be in the subspace. If I multiply it by 3 I would get the matrix [3 0 0 ; 0 3 0 ; 0 0 3] wich is clearly not in the row reduced echelon form and consequently not in the S "subspace".
Am I missing something?

anonymous · Answer

The constraint is on the "vectors" that are used to generate, or span, the subspace and not on the subspace itself.

anonymous · Answer

What you have proved is that set of row reduced echelon form matrices does not form a space. However, all linear combinations of row reduced echelon matrices, including [3 0 0; 0 3 0; 0 0 3],  is in the set of upper triangular matrices.