If a vector $v$ in $\mathbb{R}^n$ is orthogonal to each vector in a basis for a subspace $W$ of $\mathbb{R}n$, then $v$ ...... . If a vector $v$ in $\mathbb{R}^n$ is orthogonal to each vector in a basis for a subspace $W$ of $\mathbb{R}^n$, then $v$ is orthogonal to every vector in $W$ .
I don't know how to prove this statement. All possible help appreciated!
 A: Let $\{w_1, \dots, w_d\}$ be a basis for $W$ ($d < n$). We know for each $w_j$, $\langle v,w_j \rangle = 0$, where $\langle\cdot,\cdot \rangle$ is our inner product. Take any vector $w \in W$. We know for some $\alpha_j \in \mathbb{R}$, $w = \sum_{j=1}^d \alpha_jw_j$. Now, $$\langle v,w \rangle = \langle v, \sum_{j=1}^d \alpha_jw_j \rangle = \sum_{j=1}^d \alpha_j \langle v,w_j \rangle$$ Do you think you can conclude from here?
A: Let $B=\{x_1,x_2,...,x_r\}$ is basis of $W$ the Each vector $w\in W$ can be written as linear combination of elements of $B$, i.e.
$$w=a_1x_1+a_2x_2+...+a_rx_r$$, $a_1,a_2,...,a_r\in F$
Taking inner product with $v$ we have,
$<w,v>=<a_1x_1+a_2x_2+...+a_rx_r,v>$
$=a_1<x_1,v>+a_2<x_2,v>+...+a_r<x_r,v>$
$=a_1.0+a_2.0+...+a_r.0=0$, which means that each vector of $W$ is orthogonal to $v$
A: Yes ! it is true.
Suppose $\{x_1,x_2....x_m\}$ is basis of Subspace $W$ then 
Choose any arbitrary element $y$ in $W$ then We can write $y$ as linear combination of basis element. So 
$$y={\alpha}_1 x_1+{\alpha}_2 x_2+......{\alpha}_m x_m$$ Now $$<{\alpha}_1 x_1+{\alpha}_2 x_2+......{\alpha}_m x_m,v>={\alpha}_1<x_1,v>+{\alpha}_2<x_2,v>+......{\alpha}_m<x_m,v>=0$$ Since $v$ is orthogonal to each basis element.
So it shows that $v$ is orthogonal to each element of subspace $W$.
