Prove that $\{v_1+u_1,\cdots,v_n+u_n\}$ form a basis if $\{v_1,\cdots,v_n\}$ is an orthonormal basis and $\sum_{j=1}^n||u_j||^2<1$

Question

Let $V$ be a finite dimensional inner product space with $\{v_1,\cdots,v_n\}$ as orthonormal basis vectors. Let $\{u_1,\cdots,u_n\}\in V$ be vectors such that $\sum_{j=1}^n||u_j||^2<1$, the prove that $\{v_1+u_1,\cdots,v_n+u_n\}$ form a basis for $V$, where $||u_j||$ is the norm of the vector $u_j$

I don't even see a clue of how the property $\sum_{j=1}^n||u_j||^2<1$ makes it a basis ?

Answer 1

Your condition should be $\sum_{i=1}^n \|u_i\|^2 < 1$. Assume $$\sum_{i=1}^n \alpha_i(v_i+u_i) = 0$$ for some scalars $\alpha_i$ and consider the scalar product $\langle \cdot, v_j\rangle$. We get $$0 = \sum_{i=1}^n \alpha_i \langle v_i+u_i, v_j\rangle = \alpha_j + \sum_{i=1}^n \alpha_i \langle u_i, v_j\rangle$$ so using Cauchy-Schwarz follows $$|\alpha_j|^2 = \left|\sum_{i=1}^n \alpha_i \langle u_i, v_j\rangle\right|^2 \le \left(\sum_{i=1}^n |\alpha_i|^2\right)\left(\sum_{i=1}^n \left|\langle u_i,v_j\rangle\right|^2\right).$$ Summing this up over $j=1, \ldots, n$ and using Parseval we get $$\sum_{j=1}^n |\alpha_j|^2 \le \left(\sum_{i=1}^n |\alpha_i|^2\right)\left(\sum_{i,j=1}^n \left|\langle u_i,v_j\rangle\right|^2\right) = \left(\sum_{i=1}^n |\alpha_i|^2\right)\underbrace{\left(\sum_{i=1}^n \|u_i\|^2\right)}_{<1}$$ which implies $\sum_{i=1}^n |\alpha_i|^2 = 0$ so all $\alpha_1 = \cdots = \alpha_n=0$.

Answer 2

I assume your condition is $\sum_j||u_j||^2<1.$ If so, then we can prove the conclusion by contradiction. Suppose $$\sum_ja_j(u_j+v_j)=0$$ for some $(a_1,\cdots,a_n)\neq 0.$ Then $$\sum_ja_j^2=||\sum_ja_jv_j||^2=||\sum_ja_ju_j||^2\le \left(\sum_ja_j^2\right)\left(\sum_j||u_j||^2\right)<\sum_ja_j^2$$ since $\sum_ja_j^2>0.$ Thus the conclusion follows.