Proving $\left|\left|\vec{w}\right|\right|^2 \ge \sum_{j=1}^n\left|\langle\vec{w},\vec{u}_j\rangle\right|^2$.

Question

Proving $\left|\left|\vec{w}\right|\right|^2 \ge \sum_{j=1}^n\left|\langle\vec{w},\vec{u}_j\rangle\right|^2$.

$\vec{u}_1,...,\vec{u}_j$ is a set of orthonormal vectors in an inner product space $V$. $\vec{w}\in V$

I'm trying to understand this before my test tomorrow morning. Any help would be appreciated.

Thank you!

Answer 1

The question is slightly wrong. If $u_1,\cdots,u_n$ is an orthonormal basis of the vector space $V$ then for any vector $w\in V$ you can write $$w=c_1u_1+\cdots+c_nu_n$$ Then $\langle w,u_j\rangle=c_j$. Hence $$w=\sum_{i=1}^n\langle w,u_i\rangle u_i$$ Then $$||w||^2=\langle w,w\rangle=\left\langle\sum_{i=1}^n\langle w,u_i\rangle u_i,w\right\rangle=\sum_{i=1}^n\langle w,u_i\rangle\langle u_i,w\rangle$$ Hence proved.