Let $W$ be a linear subspace of $V$ such that $V = \ker(f)⊕W$ where $\ker(f)$ is the null space of $f$. What is the dimension of $W$? Let $V$ be a vector space (over $ \mathbb{R}$) of dimension 7 and let $f : V → \mathbb{R}$
be a non-zero linear functional. Let $W$ be a linear subspace of $V$ such that
$V = \ker(f)⊕W$ where $\ker(f)$ is the null space of $f$. What is the dimension
of $W$?
 A: $W$ has dimension $1$. Use the rank-nullity theorem and the fact that $f$ is non-zero to conclude that the rank is $1$ and $Ker(f)$ has dimension $6$.
A: By rank-nullity, $\dim V  -  \dim \ker f = 1 = \dim W$.  Notice that the dimension of $V$ is not relevant.
A: Since $\dim V=\dim(\ker f)+\dim W$ it follows due to the Rank-Nullity theorem that, $\dim\Im(f)=\dim W.$
$\Im(f)$ is a subspace of $\mathbb R.$ Naturally enough $\dim\Im(f)\le1.$
Now $f\ne0\implies\dim\Im(f)\ne0.$ Consequently, $\dim W=1.$
A: You can do this even without invoking rank-nullity. If one restricts $f$ to $W$, this gives a linear map $W\to\Bbb R$ that is nonzero (for if it were then $f$ would be zero on all of $\ker(f)+ W=V$, which it isn't) and therefore surjective (since $\Bbb R$ has dimension$~1$ only); moreover the kernel of the map is $\ker(f)\cap W=\{0\}$, so it is injective. The restriction is therefore an isomorphism, and its existence implies $\dim(W)=\dim(\Bbb R)=1$.
In this argument the dimension of$~V$ does not enter at all. So the conclusion remains valid for $V$ of any dimension, even infinite.
