Tupel of lineary independent linear forms gives a surjection

Question

Let $V$ be a finite dimensional $K$-vector space of dimension $n$ and $(\phi_1,...,\phi_r)$ lineary independent elements of $V^\ast$.

Considering the map $f=(\phi_1,...,\phi_r):V\to K^r$, the kernel of $f$ turns out to be the intersection of the kernels of the $\phi_i$'s. But is $f$ surjektive?

Answer 1

Yes. By the rank-nullity theorem$$\dim\operatorname{Im}F=n-\dim\ker f=n-(n-r)=r=\dim K^r.$$Therefore, $\operatorname{Im}F=K^r$.

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Here's another proof. Since $V$ has dimension $n$, you can assume without loss of generality that $V=K^n$. Suppose that $f$ is not surjective. Then there are scalars $a_1,\ldots,a_n$, not all of which are $0$, such that$$\operatorname{Im}F\subset\left\{(x_1,\ldots,x_n)\in K^r\,\middle|\,\sum_{k=1}^ra_kx_k=0\right\},$$since $\operatorname{Im}F\neq K^r$ and therefore $\operatorname{Im}F$ is contained in some hyperplane. But then$$\bigl(\forall(x_1,\ldots,x_n)\in K^r\bigr):\sum_{k=1}^ra_k\phi_k(x_1,\ldots,x_n)=0.$$In other words, $\sum_{k=1}^ra_k\phi_k=0$, which is impossible, since the $\phi_k$'s are linearly independent.

Answer 2

$f$ is surjective, since given $(x_1,\dots,x_r)\in K^r$, take $a_i=\frac {x_i}{\phi_i(1)}$. Then $f(a_1,\dots,a_r)=(a_1\phi_1(1),\dots,a_r\phi_r(1))=(x_1,\dots, x_r)$. Note: $\phi_i(1)\not=0 \;\forall i$, by linear independence...