A basis for M2x2 would have 4 vectors/matrices, but how many would be in a basis for H?judahs_lion said:H = ([a,b;c,d] : a+d =0}
Dim(M2x2)= 4, so a basis would have 4 components?
judahs_lion said:I got this far and am stuck.
[a, b ; c , -a] = a[1,0; 0,-1] + b[0, 1;0,0] +c[0,0; 1,0]
For H to be a subspace of M2x2, it must satisfy three conditions: it must contain the zero vector, it must be closed under vector addition, and it must be closed under scalar multiplication. This means that any vector in H must also be in M2x2, and the sum of any two vectors in H must also be in H. Additionally, multiplying any vector in H by a scalar must result in another vector in H.
To prove that H is a subspace of M2x2, you must show that it satisfies all three conditions mentioned above. You can do this by taking an arbitrary vector in H and showing that it is also in M2x2, as well as showing that the sum of any two vectors in H is also in H and that multiplying any vector in H by a scalar results in another vector in H.
No, the zero vector is a necessary component of a subspace. Without it, the subspace would not be closed under scalar multiplication, as multiplying any vector by 0 would result in a vector outside of the subspace.
No, a subspace must be closed under vector addition in order for it to be a valid subspace. If the sum of two vectors is not in H, then H would not satisfy the condition of being closed under vector addition.
A subset is a collection of elements that are part of a larger set. A subspace, on the other hand, is a subset that also satisfies certain conditions, such as being closed under vector addition and scalar multiplication. In other words, all subspaces are subsets, but not all subsets are subspaces.