If I have $f : V^* \rightarrow \mathbb{R}$ in the dual space of $V$, with $dim(V) = n$. I read somewhere that $dim(ker(f)) \ge n-1$, why is this the case? If I take $\{v_1,v_2,…,v_n\}$ to be a basis of $V$ and $\{f_1,…,f_n\}$ to be the corresponding dual basis of $V^*$.
Surely if I take $f = \sum_{i=1}^n f_i$, I have the vectors $v_1,…,v_n$ are not in the Kernel individually, surely that contradicts above!
Best Answer
Let
$$V = \mathbb{R}^3 = \text{span}\left\{\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} \right\}.$$
The corresponding dual basis for $V^*$ can be written as the span of row vectors (covectors act on vectors the same way row vectors act on column vectors). We can write,
$$V^* = \text{span}\left\{\begin{bmatrix} 1 & 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0 & 1\end{bmatrix} \right\}.$$
Define,
$$f = \begin{bmatrix} 1 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 1 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 1 & 1\end{bmatrix}.$$
It is apparent that $f$ has a kernel of dimension 2 spanned by,
$$\begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix},\quad \begin{bmatrix} 0 \\ 1 \\ -1\end{bmatrix}.$$
The mistake in your reasoning is that you tried to argue that the kernel of the sum of linear maps is just the intersection of the kernel of those linear maps. This is not true.
$$\text{ker}\left(\sum_{i=1}^3 f_i\right) \neq \bigcap_{i=1}^3 \ker(f_i).$$