Proof explanation: Linear dependence of functionals.

Question

Good night, i have a serious doubts with this proof. If someone can explain me the red parts i will be very grateful.

Answer 1

If $$\tag{1}\bigcap_{j\ne k}\ker f_j=\bigcap_j\ker f_j,$$ this means that $$ \bigcap_{j\ne k}\ker f_j=\ker f_k\cap\bigcap_{j\ne k}\ker f_j, $$ so $\bigcap_{j\ne k}\ker f_j\subset\ker f_k$. This allows us to restart the argument of the Lemma and prove that $f_k$ is a linear combination of $\{f_j\}_{j\ne k}$ (we may need to recycle the argument down if the intersection still doesn't have the property, but with one functional less on each step, eventually condition $(1)$ fails, or we run out of functionals); then it is enough to prove the Lemma for $\{f_j\}_{j\ne k}$.

Your second question: $$ f_k(x_k)=f_k\left(\frac{y_k}{f_k(y_k)}\right)=\frac{f_k(y_k)}{f_k(y_k)}=1. $$

You have, by linearity $$ f_j(y)=f_j\left(x-\sum_{k=1}^n f_k(x)x_k\right) =f_j(x)-\sum_{k=1}^n f_k(x)f_j(x_k)=f_j(x_k)-f_j(x)f_j(x_j)=0, $$ the second to last equality because $f_j(x_k)=0$ for $j\ne k$, and the last one because $f_j(x_j)=1$.