Let $V$ be a linear space s.t. $\dim\, V=n$ and $f_1,f_2,\dots,f_m$ be linear functionals ($m<n$). Show that $\exists x\neq 0$ s.t. $f_i(x)=0$ $\forall i$.

We can suppose that these $f_i$ are linearly indepent.

Let $\{e_i\}_{i=1}^n$ be a basis to $V$ and $\{\delta_i\}_{i=1}^n$ be its dual basis.

We can also set $f_i=\sum_{k=1}^n\alpha_{ik}\delta_i$. Then, $f_i(e_j)=\alpha_{ij}$

Let $x\in V$, then $x=\sum_{j=1}^n \beta_j e_j$.

Therefore, we have that $f_i(x)=\sum_{j=1}^n\beta_j\alpha_{ij}$.

And I'm stuck here.

  • $\begingroup$ All the $f_i$ map $V\to F$ where $F$ is the field of the vector space $V$? E.g. $\mathbb{R}$? ("functional" generally means that). $\endgroup$ – BrianO Nov 8 '15 at 19:16
  • $\begingroup$ @BrianO yes, they're linear functionals. $\endgroup$ – Andre Gomes Nov 8 '15 at 19:18
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    $\begingroup$ math.stackexchange.com/questions/1089498/… $\endgroup$ – R.N Nov 8 '15 at 19:22

$F = (f_1,\ldots,f_m)$ defined by $F(x) = (f_1(x),\ldots,f_m(x))$ is a linear map from $V$ to $F^m$.

$\dim(\ker(F)) + \dim(\operatorname{im}(F)) = \dim(V)$. We know that the right hand side equals $n$. The image of $F$ has dimension at most $m < n$. So the dimension of $\ker(F)$ cannot be $0$, so has some vector $x \neq 0$ in it. This $x$ works.

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Consider the linear transformation $f$: $$ f\colon x\mapsto (f_i(x)\mid 1 \le i \le m)\colon V\to F^m $$ where $F$ is the field of the vector space $V$.

We have $$ \dim (\ker (f)) + \dim( \operatorname{image} (f)) = \dim (V) = n. $$ But $\dim( \operatorname{image} (f)) \le m < n$, so we must have $\dim (\ker (f)) > 0$.

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