In a $d$-dimensional subspace of $\mathbb{R}^n$ there is a vector perpendicular to $d - 1$ orthogonal vectors Let $\mathcal{S}$ be a subspace of $\mathbb{R}^n$ with $\boldsymbol{s_1},\dotsc,\boldsymbol{s_m}$ as orthogonal basis vectors and let $\boldsymbol{x_i},\dotsc,\boldsymbol{x_{m - 1}}$ be orthogonal vectors in $\mathbb{R}^n$. This article on the SVD implies in the proof on the fifth page, that there exists a vector $\boldsymbol{a} \in \mathcal{S}$ that is perpendicular to all of $\boldsymbol{x_i},\dotsc,\boldsymbol{x_{m - 1}}$. Prove this, please.
 A: We put $E=Vect\{(x_j)_{i\leq j\leq m-1}\}$ $E$ is a subspace of $\mathbb{R}^n$, and the $\dim(E)<\dim(\mathbb{R}^n)=n$
So $E$ has a supplementary subspace $F$ i.e $E\oplus F=\mathbb{R}^n$,Let $(y_k)$ a basis of $F$ so by definition $\{(x_j)_{i\leq j\leq m-1}\}\cup \{(y_k)\}$ is a basis of $\mathbb{R}^n$ by gram-schmidt process we can construct an orthogonal basis $(f_l)$from $\{(x_j)_{i\leq j\leq m-1}\}\cup \{(y_k)\}$ that verifie :
$$
vect((f_l)_{i\leq l \leq m-1})=E
$$
and 
$$
<f_m,x>=0 \qquad \forall x\in E
$$
witch $a=f_m$ complete the proof.
A: Because the vectors $x_j$ are non-zero and mutually orthogonal, then every vector $x$ that can be written as a linear combination
$$
        x = \alpha_1 x_1 + \alpha_2 x_2 + \cdots +\alpha_{m-1}x_{m-1}
$$
must be equal to
$$
                  x=\frac{(x,x_1)}{(x_1,x_1)}x_1+\frac{(x,x_2)}{(x_2,x_2)}x_2+\cdots+\frac{(x,x_{m-1})}{(x_{m-1},x_{m-1})}x_{m-1}.
$$
However, there must be a non-zero vector that cannot be so written because the dimension of the space is $m$. For this $x$, the vector $a$ shown below is non-zero:
$$
       a=x-\frac{(x,x_1)}{(x_1,x_1)}x_1-\frac{(x,x_2)}{(x_2,x_2)}x_2-\cdots-\frac{(x,x_{m-1})}{(x_{m-1},x_{m-1})}x_{m-1}
$$
You can directly check that $(a,x_1)=(a,x_2)=\cdots=(a,x_{m-1})=0$.
