linear map to rationals I have a problem that can be solved if I use the follwing lemma.
Let $V$ be the span of $1, x_1, x_2, . . . , x_n$ over $Q$ where $x_i$ are irrationals. Then there exists $Q$-linear mapping $f$ : $V$ → $Q$ such
that $f (1) = 0$ and $f (x_i) \neq 0$.
but is it true? If it is how can I prove it?
 A: Here's an idea:
Let $x_0 = 1$.  We may select a subset $\mathcal B = \{y_0,y_1,\dots,y_{d-1}\} \subset \{x_0, x_1,\dots,x_n\}$ such that $y_0 = x_0 = 1$ and $\mathcal B$ is a basis for the span of $V$. It follows that we may define a $\Bbb Q$-linear map by extending the definition
$$
f(y_i) = \begin{cases}
0 & i=0\\
a_i & \text{otherwise}
\end{cases}
$$
now, it remains to be seen that we can select $a_i$ such that $f(x_i) \neq 0$ for all $i$, which is admittedly the tricky point here.
That is, we have reduced the problem to the following claim:

Let $y_1,\dots,y_d$ be a basis of a subspace $V \subset \Bbb R$ over $\Bbb Q$.  Let $x_1,x_2,\dots,x_k \in V$ be arbitrary but non-zero.  Then there exists a functional $f$ such that $f(x_i) \neq 0$ for all $i$. 

Or equivalently,

Let $b_1,b_2,\dots,b_k$ denote non-zero elements of $\Bbb Q^d$, where $b_i = (b_{i1},\dots,b_{id})$.  Let $B$ denote the matrix whose $i$th column is $b_i$.  Then there exists an $a \in \Bbb Q^d$ such that all entries of $a^TB$ are non-zero.

Or more concisely,

Let $M$ be a (rational) matrix such that all rows of $M$ are non-zero.  Then, there exists a vector $a$ such that $Ma$ has all non-zero entries.

Sounds reasonable, but I'm not sure how to prove it.
