Linear independence of a set of vectors + orthonormal basis I have a set of vectors $\{ y_1, y_2, \dots, y_n \}$ and an orthonormal basis $\{ u_1, u_2, \dots, u_n \}$ and I know that 
$$\sum_{i=1}^n \| y_i \|^2 < 1$$
I need to prove that the set 
$$\{ u_1+y_1, u_2+y_2 ,\dots , u_n+y_n \}$$
 is linear independent. I tried doing something with the angle between the vectors, but doesn't lead anywhere and I'm not sure how I can prove this.
 A: Suppose $\{u_{1}+y_{1},...,u_{n}+y_{n}\}$ is linearly dependent. Then we can find scalar $\lambda_{1},...,\lambda_{n}\in\mathbb{R}$ such that
$$\sum_{i=1}^{n}\lambda_{i}u_{i}+\lambda_{i}y_{i}=0$$
and $\sqrt{\sum^{n}_{i=1}|\lambda_{i}|}<1$. Note that this implies that $$\sqrt{\sum^{n}_{i=1}|\lambda_{i}|}>\sum^{n}_{i=1}|\lambda_{i}|.$$
We find that
$$0=\|\sum_{i=1}^{n}\lambda_{i}u_{i}+\lambda_{i}y_{i}\|\geq\|\sum_{i=1}^{n}\lambda_{i}u_{i}\|-\sum_{i=1}^{n}|\lambda_{i}|\|y_{i}\|=\sqrt{\sum^{n}_{i=1}|\lambda_{i}|}-\sum^{n}_{i=1}|\lambda_{i}\|y_{i}\|.$$
As $\sum^{n}_{i=1}\|y_{i}\|^{2}<1$ we find that $\|y_{i}\|<1$ for all $i$, hence
$$0=\|\sum_{i=1}^{n}\lambda_{i}u_{i}+\lambda_{i}y_{i}\|\geq \sqrt{\sum^{n}_{i=1}|\lambda_{i}|}-\sum^{n}_{i=1}|\lambda_{i}\|y_{i}\|>\sqrt{\sum^{n}_{i=1}|\lambda_{i}|}-\sum^{n}_{i=1}|\lambda_{i}|=0$$
which is a contradiction.
A: We may assume that the underlying inner product space is $\mathbb C^n$ and the inner product is given by $\langle u,v\rangle=v^\ast Pu$ for some positive definite matrix $P$.
Let $U$ be the matrix whose columns are the $u_i$s. Define $Y$ analogously. Since the $u_i$s are orthonormal with respect to the inner product $\langle\cdot,\cdot\rangle$, we have $U^\ast PU=I$. Now suppose $(U+Y)v=0$. Then $-Uv=Yv$ and hence
$$
v^\ast v=v^\ast U^\ast PUv=\|-Uv\|^2=\|Yv\|^2=v^\ast Y^\ast PYv\le\lambda_\max(Y^\ast PY)v^\ast v.\tag{1}
$$
However, as $Y^\ast PY$ is positive semidefinite, $\lambda_\max(Y^\ast PY)\le\operatorname{tr}(Y^\ast PY)=\sum_i\|y_i\|^2<1$. Therefore $v$ must be zero in $(1)$, i.e. $U+Y$ is non-singular.
