From Serge Lang's Linear Algebra:
(1) Let $V$ be a vector space of finite dimension $n$ over the field $K$. Let $\phi$
be a functional on $V$, and assume $\phi \neq 0$. What is the dimension of the
kernel of $\phi$? Proof?(2) Let $V$ be a vector space of dimension $n$ over the field $K$. Let $\psi$, $\phi$ be two non-zero functionals on $V$. Assume that
there is no element $C \in K$, $c \neq 0$ such that $\psi = c \phi$.
Show that:$$\textrm{Ker} \, \phi \, \cap \, \textrm{Ker} \, \psi$$
Both of these questions seem to be correlated, so I decided to put them together.
Note: I've seen this question and there are few very interesting solutions (even in question, using Riesz-representation theorem but limited to Hilbert spaces), but in this case I wonder how valid my approach is.
My attempts
(1) We know that $\phi: K^n \rightarrow K \neq 0$ where $V = K^n$. By the rank-nullity theorem, it can be observed that:
$$\textrm{dim} \, V = \textrm{dim} \, \textrm{Im}(\phi) + \textrm{dim} \, \textrm{Ker}(\phi)$$
We also know that $\textrm{dim} \, \textrm{Im}(\phi) \leq \textrm{dim} \, K$. I will make a very important assumption here:
$$\phi \neq 0 \implies \textrm{Im}(\phi) \neq 0$$
Thus by the assumption above, $\textrm{dim} \, \textrm{Im}(\phi) = \textrm{dim} \, K = 1$ (implying surjectivity), going back to rank-nullity:
$$\textrm{dim} \, V = 1 + \textrm{dim} \, \textrm{Ker}(\phi)$$
$$\textrm{dim} \, V – 1 = \textrm{dim} \, \textrm{Ker}(\phi)$$
$$\textrm{dim} \, \textrm{Ker}(\phi) = n – 1$$
Is my assumption valid?
(2)
We see that:
$$\psi: K^n \rightarrow K \neq 0$$
$$\phi: K^n \rightarrow K \neq 0$$
The question (2) assumes that:
$$\psi \neq c \phi, \forall c \in K$$ where $c \neq 0$.
From the solution above, it can be concluded that:
$$\textrm{dim} \, \textrm{Ker}(\psi)=\textrm{dim} \, \textrm{Ker}(\phi)=n-1$$
If $c \phi \neq 0$ and $c \neq 0$, then $\phi \neq 0$, thus I make assumption that $\phi(v) \neq 0$ and therefore considering that:
$$\textrm{Ker}(\phi) = \{v \mid \phi(v) = 0 \} = \emptyset$$
we can conclude that $\textrm{dim} \, \textrm{Ker}(\phi)=0$, and then apply formula:
$$\textrm{dim}(\textrm{Ker} \, \phi \, \cap \, \textrm{Ker} \, \psi) = \textrm{dim} \, \textrm{Ker} \, \phi + \textrm{dim} \, \textrm{Ker} \, \psi – (\textrm{dim} \, \textrm{Ker} \, \phi – \textrm{dim} \, \textrm{Ker} \, \psi)$$
$$\textrm{dim}(\textrm{Ker} \, \phi \, \cap \, \textrm{Ker} \, \psi) = 0 + (n – n) = 0$$
which makes no sense, because $n – 2$ is not the result..
Question:
I'm trying to prove the question (2) with my solution, but I use the result from (1) to do so.
If result from (1) is correct, what's the problem with my solution for (2), does it make any sense at all? Can it be modified to be correct, or should I stick with the solution from the referenced question?
Thank you!
Best Answer
(1) is correct.
For (2), observe that $\dim\ker\phi=\dim\ker\psi=n-1$ by (1), so $\ker\phi=0$ is nonsense.
The dimension formula for the intersection and generatum (=sum) of subspaces correctly goes as: $$\dim(A\cap B)+\dim(A+B)=\dim A+\dim B$$ for arbitrary subspaces $A, B$.
By the conditions, it is possible to prove $\ker\phi+\ker\psi=V$, whence $$\dim(\ker\phi\cap\ker\psi)=n-2$$