How to prove the linear independence of a set of linear combinations of orthogonal vectors Given a set of $m$ orthogonal vectors $\{\phi_1, \dots, \phi_m\}$
show that if you form any $n>m$ linear combinations of them :
$$
\begin{align}
v_1 & = a_{11} \phi_1+ \dots+ a_{1m} \phi_m \\
v_2 & = a_{21} \phi_1+ \dots+ a_{2m} \phi_m \\
& \vdots \\
v_n & = a_{n1} \phi_1+ \dots+ a_{nm} \phi_m
\end{align}
$$
($a_{ij}$ coefficients arbitrary), 
then $\{v_1,\dots,v_n\}$ can not be a linearly independent set.
 A: Let $W=\operatorname{span}(\phi_1, \dots, \phi_m)$ and $U=\operatorname{span}(v_1,\dots,v_n)$. $\dim W=m$ because the vectors are orthogonal, and $U \subseteq W$. This implies that $\dim U\leq \dim W$. But $n > m$, so the vectors cannot be linearly independent.
A: Here is an approach. Assume for simplicity they are orthonormal ($\langle \phi_i,\phi_j\rangle =\delta_{ij}$). Now, we follow the steps 
step 1: construct the vectors

$$ v_k = \sum_{i=1}^{m}a_{k i} \phi_i \quad k=1,2,\dots n,  $$

which are the linear combinations of the given orthonormal set.
step 2: we make the linear combinations of the vectors $v_k's,\quad k=1,\dots,n$ and set it equal to zero

$$ \sum_{j=1}^{n}b_j v_j =0 \implies  \sum_{j=1}^{n}b_j \sum_{i=1}^{m}a_{j i} \phi_i=0. $$

step 3: Take the dot product of the above equation w.r.t. $\phi_s,\quad s=1,\dots,m $ as
$$ \langle \sum_{j=1}^{n}b_j \sum_{i=1}^{m}a_{j i} \phi_i, \phi_s \rangle   = 0, \quad s=1,2,\dots,m.  $$
$$ \implies \sum_{j=1}^{n}\sum_{i=1}^{m}b_j\,a_{j i} \delta_{is}=0,\quad s=1,\dots,m. $$
$$ \implies \sum_{j=1}^{n} b_j\,a_{j s} \delta_{ss}=0, \quad s=1,\dots,m. $$

$$ \implies \sum_{j=1}^{n} b_j\,a_{j s} = 0, \quad s=1,\dots,m \longrightarrow (*)$$

Now, one can see that $(*)$ is an underdetermined system of equations in $b'_js$, since $n>m$ which means we will have $n-m$ free variables and the system will have infinite number of solutions.
