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!
Best Answer
You were wrong because your intuition misled you. I don't mean to offend; it happens to everyone who studies math. Your intuition told you that the set of vectors perpendicular to $v$ would be a plane. This is because you are used to imagining with vectors in $\mathbb R^3$. In $\mathbb R^3$, we can choose only three independent directions at a time, and all other directions are some combination of those. You have one direction you cannot go, so that leaves two. But in $\mathbb R^4$, you have four directions! You cannot go in the direction of $v$, but you still have three other directions. In general, if you are looking for the orthogonal complement of a $k$ dimensional subspace of $\mathbb R^n$, it will have $n-k$ dimensions.