Let $V$ be a vector space with inner product $\langle\,,\,\rangle$ and $\{v_1,v_2,\dots,v_n\}$ an orthonormal basis of $V$. Prove that $\sum_{i=1}^n \langle x,v_i\rangle^2=\|x\|^2$, for $x\in V$.

I´m not very fond of proofs but this is what I have:

\begin{align} \sum_{i=1}^n \langle x,v_i\rangle^2 &=\sum_{i=1}^n (\|x\|\,\|v_i\| \cos(\theta))^2\\ &=\sum_{i=1}^n \|x\|^2 \|v_i\|^2 \cos^2(\theta)\\ &=\sum_{i=1}^n (\|x\|^2\cos(\theta)) (\|v_i\|^2 \cos(\theta))\\ &=\sum_{i=1}^n \langle x,x\rangle \langle v_i,v_i\rangle\\ &=\langle x,x\rangle\sum_{i=1}^n \langle v_i,v_i\rangle\\ &=\|x\|^2 n \end{align}

What am I wrong? Any suggestions to improve my proofs would be helpful. Thanks.

  • $\begingroup$ You are wrong from the beginning, as the angle $\theta$ depends on $i$. $\endgroup$ – Yves Daoust Dec 20 '16 at 23:09

Your mistake is this: $$\|x\|^2\cos\theta \ \ne\ \langle x,x\rangle$$ Instead, we have $\langle x,x\rangle = \|x\|^2$ without the $\cos$ term.

And, hint for the solution: write $x=\xi_1 v_1+\xi_2 v_2+\dots$, then calculate $\|x\|^2=\langle x,x\rangle$ by substituting this into the right argument.


This is known as the Parseval's Identity. Observe that since $(v_i)_{i=1}^n$ is an orthonormal basis, $$ x = \sum_{i=1}^n \langle v_i,x \rangle v_i $$ By Pythagorean Theorem, $$ \|x\|^2 = \left\lVert \sum_{i=1}^n \langle v_i,x \rangle v_i\right\rVert = \sum_{i=1}^n \|\langle v_i,x \rangle v_i\|^2 = \sum_{i=1}^n |\langle v_i,x \rangle|^2 $$ The problem with your proof is it uses $\theta$, which is not very rigorous, and leads to the incorrect result.


$$x=\sum_{i=1}^n v_ix_i,$$ where $x_i=\langle x,v_i\rangle$ are scalars.

Indeed, by linearity and orthonormality, $$\langle x,v_j\rangle=\sum_{i=1}^n \langle v_j,v_i\rangle x_i=\sum_{i=1}^n \delta_{ij} x_i=x_j.$$


$$\|x\|^2=x^2=\sum_{i=1}^n\sum_{j=1}^n \langle v_i,v_j\rangle x_i x_j=\sum_{i=1}^n\sum_{j=1}^n \delta_{ij} x_i x_j=\sum_{i=1}^n x_i^2.$$


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