Question

Fine the orthogonal complement of the subspace
SPAN ((1, 1, 1, 1,), (1, 1, -1, -1) in R^4 with the dot product as the inner product

Answer 1

How about the space spanned by the vectors
\[<1,-1,-1,1> , <1,-1,1,-1>\]

Answer 2

how did you determine that?

Answer 3

when there are a lot of 1's positive and negative, you can kinda just guess and check. If you dont want to do that, theres always the Gram-Schmidt process.

Answer 4

how would you go about doing the gram-schmidt process?

Answer 5

Eesh, please don't try to do Gram-Schmidt for something like this. It would consist of arbitrarily writing down vectors that are linearly independent from those two and then stripping them down to force them to be orthogonal. This is a case when it's pretty easy to just look and see two vectors that are orthogonal to the ones given, but you could also just make an arbitrary vector
\[<a,b,c,d>\]
And require that its dot product with the two given vectors be zero. From the first dot product,
\[a+b+c+d = 0\]
and from the second,
\[a+b-c -d = 0\]
So apparently
\[b = -a\]
and
\[d = -c\]
so your arbitrary vector becomes
\[a,-a,c,-c\]
now you can just select whichever values you please. I am partial to ones, so if
\[a = c = 1\]
we have
\[<1,-1,1,-1>\]
and if a and c have different signs, we get
\[1,-1,-1,1>\]
But you see you have quite a few choices, you just need to make sure that the two new ones you choose are not linearly dependent.