I need help with showing this:
If f is a linear functional on an n-dimensional vector space X, what dimension can the Null Space N(f) have?
Answer:
dim (N(f)) = n or n-1.
If N(f) = X then its dimension is n. Otherwise, there must be some element which is not in the null space of f, and by taking a scalar multiple of it, we may assume that . If x is any element of X then , and . So every element of X is a sum of something in the 1-dimensional space spanned by and something in N(f). You should be able to deduce from this that dim(N(f)) = n–1.
Hi Opalg, thank you so much for your reply.
Unfortunately I dont follow the whole argument yet.
Can you help me a bit more please?
I follow first part, then...
So every element of X is a sum of something in the 1-dimensional space spanned by and something in N(f). You should be able to deduce from this that dim(N(f)) = n–1.....
I intuitively can see that dim(span{x_0})=1 and N(f) must then be of dimension n-1 since dimX=n. Is there a mathematical way to show this using n-1 linear independent vectors?