# Thread: Dimension of the Null Space of Linear Functional

1. ## Dimension of the Null Space of Linear Functional

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?

dim (N(f)) = n or n-1.

2. Originally Posted by frater_cp
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?

dim (N(f)) = n or n-1.
If N(f) = X then its dimension is n. Otherwise, there must be some element $x_0\in X$ which is not in the null space of f, and by taking a scalar multiple of it, we may assume that $f(x_0)=1$. If x is any element of X then $x = f(x)x_0 + (x-f(x)x_0)$, and $x-f(x)x_0\in N(f)$. So every element of X is a sum of something in the 1-dimensional space spanned by $x_0$ and something in N(f). You should be able to deduce from this that dim(N(f)) = n–1.

3. ## Full of Gratitude

Unfortunately I dont follow the whole argument yet.
Can you help me a bit more please?

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?

4. Let $\{x_1,x_2,\ldots,x_k\}$ be a basis for N(f) (so that k is the dimension of n(f), which we don't yet know). Then the calculation in my previous comment shows that $\{x_0,x_1,x_2,\ldots,x_k\}$ is a basis for the whole of X, and therefore contains n vectors. So k+1=n, or k=n–1.

5. ## Gratitude

excellent excellent excellently done!!

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# null space of linear functional

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