Dimension of Basis of Subspace I was just wondering if there is any "rule" for what the dimension of a basis of a given subspace will be.
For example, a problem I just did involved a vector $v = (1, 2, 3, 4)$ in $\mathbb{R}^4$, and I had to find a basis for the subspace in $\mathbb{R}^4$ consisting of all vectors perpendicular to $v$.
My intuition for this was to note that the subspace of vectors perpendicular to v is the plane with v as its normal vector. Thus, any two vectors in the plane which are linearly independent would be a basis, and the dimension of the basis would be two.
However, the answer the book gave had a dimension of three. They solved it in a way that makes sense to me as well but I'm still confused as to where I went wrong with my logic. I'm sensing the answer lies in the fact that R4 planes don't behave the same way as R3 planes.
Another problem involved finding the basis for the orthogonal complement of a subspace in R4 spanned by two vectors: $W = Span\{(1, 2, 3, 4),(5, 6, 7, 8)\}$, and the answer had a dimension of two (not three).
I guess I'm just confused on where the dimension is coming from. Any help would be great.
Thanks!


My intuition for this was to note that the subspace of vectors perpendicular to v is the plane with v as its normal vector. Thus, any two vectors in the plane which are linearly independent would be a basis, and the dimension of the basis would be two.

Unfortunately planes are not always two-dimensional. The correct intuition is that the orthogonal complement of a subspace $U$ has dimension that "complements" the dimension of $U$ (assuming everything is finite-dimensional). So in a 5-dimensional space, the orthogonal complement of a one-dimensional subspace is 4-dimensional, the orthogonal complement of a two-dimensional subspace is 3-dimensional, and so forth.
Until you build a strong geometric intuition for vectors in dimensions higher than 3, I would suggest going with the direct approach of forming a basis and counting the number of elements.
