I have the following:
(Note: $V^{*}$ is defined as: $V^{*} = \{ L: V \rightarrow \mathbb{R} | \text{L is linear} \}$)
Let $V$ be an $\mathbb{R}$-Vectorspace. Let $\phi \in V^{*} \text{ \ } \{0 \}$ and $h \in V\text{ \ } \{0 \}$ with $h \in \ker{\phi}$. Let now $F: V \rightarrow V$ be given as $F(v) := v + \phi(v) h$, where $B = (h, v_2, v_3 ... v_n)$ is a basis of $V$ with its first basis-vector $h$. Calculate the Transformation-Matrix $M_B(F)$.
Well, what I did was the trivial. I've applied $F(b_j) = \sum_{i = 1}^n \lambda_i e_i$ with $b_j \in B$ and $j = 1,2,3 ... $
I will present the first two; $F(b_1)$ and $F(b_2)$. Note that $h\in ker(\phi) \Rightarrow \phi(h) = 0$:
$$f(h) = h + \phi(h)h = h = \lambda_1 e_1 + \lambda_2 e_2 + .. + \lambda_n e_n$$
$$f(v_2) = v_2 + \phi(v_2)h = \lambda_1 e_1 + \lambda_2 e_2 + .. + \lambda_n e_n$$
So what do I do next? I don't know how I can calculate the $\lambda$'s to be able to put them as entries into the matrix...
What if I said $h = \left(h_1, h_2, h_3, .... ,h_n \right)$ and defined every other vector $v_i = (v_{i,1}, v_{i,2}, ... ,v_{i,n})$.
Then I could write for $h$, that $h = h_1e_1 + h_2e_2 + h_3e_3 ... h_n e_n$ and for $v_2 + \phi(v_2)h = v_{2,1} + \phi(v_{2,1})h_1 e_1 + v_{2,n} + \phi(v_{2,2})h_2 e_2 + ... + v_{2,n} + \phi(v_{2,n})h_n e_n $...
Is that possible?
Thank you very much for your help.
FunkyPeanut