Question

Find a basis for the subspace of R^4 containing all the vectors whose components add up to 0.

Answer 1

I think it is the nullspace of the matrix

[1 1 1 1]

Answer 2

Basis would then be

(-1, 0, 0, 1)
(-1, 0, 1, 0)
(-1, 1, 0, 0)

Answer 3

It also asks to find a basis of the sobspace containing all the vectors that are perpendicular to the above subspace

Answer 4

No, they will be (1,−1,0,0), (1,0,−1,0), and (1,0,0,−1).

Answer 5

How is your basis any different from mine ?

Answer 6

There are no spaces after the comma XXXX

Answer 7

As long as they are linearly independent and span the subspace, you're fine

Answer 8

The bases seem pretty much the exact same thing to me, just differ by a scalar multiple, -1.

Since this is a 3D subspace of 4D space, there can only be a single vector in the space perpendicular to them. One way you could find it is by taking the determinant of these 3 vectors with 4 unit vectors like you would for a cross product that you'd normally think of in 3D space.

Alternatively, I know there's some other ways but it's not coming to me off hand but shouldn't be too difficult I just gotta remember what it is lol.

Answer 9

Oh right
Since its dimension is 1, can i simply take the basis to be the row vector (1, 1, 1, 1) ?

Answer 10

Yup any scalar multiple of \(\begin{bmatrix}1&1&1&1\end{bmatrix}^\intercal\) would suffice for the basis of the nullspace.