Dual basis and annihilator problem I think they're fairly simple but I really don't know where to start, the problems are these:
First one: 
$V$ a vector space of dimension $n$ and $\phi \in V^* \setminus \{0\}$. 
Prove that $\dim \ker \phi = n-1$.
Second one:
Let $B = \begin{bmatrix}2 & -2 \\ -1 & 1\end{bmatrix}$ and $W = \{A \in \operatorname{Mat}_{\mathbb R}(2\times2) : AB = 0\}$. Suppose $f \in W^o$ (the annihilator of $W$) such that $f(\operatorname{Id}_{2\times2}) = 0$ and $f\left({\scriptstyle\begin{bmatrix}0 & 0 \\ 0 &1\end{bmatrix}}\right) = 3$. Calculate $f(B)$.
For this is one I don't really know when I have to use that $AB = 0$.
 A: For the first question, recall the Rank-Nullity Theorem: for any linear operator $T:V\to W$ between two finite dimensional vector spaces the following equality holds: $\dim\text{ker}\ T+\dim\text{Im}\ T=\dim V$. If $\phi\in V^*\setminus\{0\}$, then $\phi:V\to\mathbb{F}$, where $\mathbb{F}$ is the field over which you consider $V$ and $W$. The dimension of $W=\mathbb{F}$ over $\mathbb{F}$ is 1. Now use the theorem.
Concerning the second question, notice that $W$ is the subspace of matrices of the form:
$$\begin{bmatrix}{} 
a & 2a\\
c & 2c
\end{bmatrix}$$
where $a,c\in\mathbb{R}$. Thus $\dim W=2$. As $f(Id)=0$, $Id\not\in W$ and $f\neq0$, we have that $\ker\ f=W\oplus\text{span}\{Id\}$. Now we have:
$$f(B)=f\begin{bmatrix}{} 
2 & -2\\
-1 & 1
\end{bmatrix}=f\begin{bmatrix}{} 
-1 & -2\\
0 & 0
\end{bmatrix}+f\begin{bmatrix}{} 
0 & 0\\
-1 & -2
\end{bmatrix}+f\begin{bmatrix}{} 
3 & 0\\
0 & 3
\end{bmatrix}$$
The first and the second matrices from RHS are in W, the third one is the multiplication of $Id$. So $f(B)=...$
