$\dim(\ker\varphi\cap\ker\psi)=n-2$ proof 
Let $V$ be a vector space of dimension $n$ over the field $K$. Let $\psi,\varphi$ be two non-zero functionals on $V$. Assume that there is no element $c\in K$,$c\neq 0$ such that $\psi=c\varphi$, Show that:
$(\ker\varphi)\:\cap\:(\ker\psi)$
has dimension $n-2$

I know $\dim V=\dim Im+\dim \ker$, in which $Im$ is image. We can apply this equation because $V$ is finite-dimensional.
$\dim(\ker\varphi)=n-1\\\dim(\ker\psi)=n-1$
If $\{v_1,v_2...v_n\}$ is a basis of V.
As we are dealing with linear functionals we can write:
$\varphi(v)=\langle v_i^{*},v\rangle$ in which $v$ stands for all vectors in $V$ and $v_i^{*}\in V^{*}$ the dual space.
$\psi(v)=\langle v_j^{*},v\rangle$ in which $v$ stands for all vectors in $V$ and $v_j^{*}\in V^{*}$ the dual space.
As $\psi=c\varphi$ for no $c\in K,c\neq 0$, so $v_i^{*}\neq v_j^{*}$
The orthogonal vectors in $V$ to $v_i^{*}$ are different from the orthogonal vectors in $V$ to $v_j^{*}$. So the statement is proven.$\blacksquare$
Questions:
1) Could someone check if my proof is right?
2) Is there any improvement recommended?
Thanks in advance!
 A: 
I know $\dim V=\dim Im+\dim \ker$

Then that's all you need.
Hint: Consider the map $T:V  \to K^2$ defined by $T(v) = (\phi(v),\psi(v))$.  Verify that $\ker(T) = \ker(\phi) \cap \ker(\psi)$.  Use the fact that there is no such constant $c$ to show that $\operatorname{Im}(T) = K^2$.
One approach is to note that
$$
\psi = c \phi \iff (c,-1) \in \operatorname{Im}(T)^\perp
$$

Proof: Consider the map $T:V  \to K^2$ defined by $T(v) = (\phi(v),\psi(v))$.  This is a linear map [show that this is true].  The kernel of $T$ is $\ker(\phi) \cap \ker(\psi)$ [show that this is true].  By the rank nullity theorem (that is, because $\dim V=\dim Im+\dim \ker$), we know that the kernel of $T$ has dimension at least $n-2$ and at most $n$.
Now, suppose for the purpose of contradiction that the dimension of the kernel of $T$ is at least $n-1$.  Then the image of $T$ has dimension at most $1$.  So, there exists a $c \in K\setminus \{0\}$ such that we can write
$$
\operatorname{Im}(T) \subseteq \{(x_1,x_2) \in K^2 : cx_1 - x_2 = 0\}
$$
(argue this part carefully!) However, this would mean that for all $v \in V$, we have 
$$
c \varphi(v) - \psi(v) = 0
$$
which would mean that $\psi = c \varphi$, contradicting our premise.
A: "If {v1,v2...vn}{v1,v2...vn} is the basis of $V$.
We know the $\ker ψ=\{v_1,v_2...v_n\}−v_a$ and $\kerφ=\{v_1,v_2...v_n\}−v_b$ ."
This sentence is false. An arbitrary basis of $V$ will not satisfy this.
Perhaps a good aproach would be using Riesz' Representation Theorem:
https://en.wikipedia.org/wiki/Riesz_representation_theorem
A: I have multiple concerns. Brace yourself!


*

*There is no "the" basis for $V$. Most vector spaces have multiple bases (in most cases, infinitely many).

*Unless you construct a basis $(v_1, \ldots, v_n)$ carefully, it may well be the case that $\phi(v_i) \neq 0$ for all $i$. Take for example, $V = \mathbb{R}^2$, $v_1 = (1, 0)$, $v_2 = (0, 1)$, $\phi(x, y) = x + y$.

*I don't see what is gained by introducing the $\langle \cdot, \cdot 
\rangle$ notation. Are you just renaming $\phi, \psi$?

*"Orthogonal" doesn't mean much in this context. Are we an an inner product space?

