4
$\begingroup$

Let $V$ be a real inner product space, and let $v_1,v_2, \dots ,v_k$ be an orthonormal set of vectors. How do you prove that

$$\sum_{i=1}^k | \langle x,v_i \rangle \langle y,v_i\rangle| \leq \|x\|\cdot\|y\|?$$

When does the equality hold?

I've been trying to do this with the Bessel and Cauchy-Schwarz inequalities, but I can't make it work yet. Any help would be greatly appreciated.

$\endgroup$
5
  • 5
    $\begingroup$ Hint: $\langle x, v_i\rangle$ are coordinates. $\endgroup$
    – cardinal
    Nov 20 '11 at 15:51
  • $\begingroup$ Use cardinal's excellent hint, and look again at Cauchy-Schwarz. $\endgroup$
    – robjohn
    Nov 20 '11 at 15:57
  • $\begingroup$ @robjohn: I'm sorry, but I am still very confused! $\endgroup$
    – Freeman
    Nov 20 '11 at 17:08
  • $\begingroup$ @cardinal: How do you calculate the coefficient of $v_i$ in the unique linear combination of $v_1,\dotsc,v_k$ which equals $v$? $\endgroup$
    – user3533
    Nov 20 '11 at 18:01
  • $\begingroup$ I have used cardinal's hint in my answer (although I don't mention it explicitly). I hope that that clears things up. $\endgroup$
    – robjohn
    Nov 20 '11 at 18:05
6
$\begingroup$

We don't necessarily know that $\{v_i:1\le i\le k\}$ is a basis of $V$ since we don't know the dimension of $V$, but we are given that $\{v_i\}$ are orthonormal; that is $\left<v_i,v_j\right>=0$ when $i\not=j$ and $\left<v_i,v_i\right>=1$.

Consider the vector $$ x^\perp=x-\sum_{i=1}^k\left<x,v_i\right>v_i\tag{1} $$ $x^\perp$ is perpendicular to $\{v_i\}$: $$ \begin{align} \left<x^\perp,v_j\right> &=\left<x-\sum_{i=1}^k\left<x,v_i\right>v_i,v_j\right>\\ &=\left<x,v_j\right>-\left<x,v_j\right>\left<v_j,v_j\right>\\ &=0\tag{2} \end{align} $$ Therefore, $x^\perp$ is perpendicular to $x-x^\perp=\sum\limits_{i=1}^k\left<x,v_i\right>v_i$.

Next consider $$ \begin{align} \left<x-x^\perp,y-y^\perp\right> &=\left<\sum_{i=1}^k\left<x,v_i\right>v_i,\sum_{j=1}^k\left<y,v_j\right>v_j\right>\\ &=\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\left<v_i,v_i\right>\\ &=\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\tag{3} \end{align} $$ Note that since $x^\perp$ is perpendicular to $x-x^\perp$, $$ \|x-x^\perp\|^2+\|x^\perp\|^2=\|x\|^2\tag{4} $$ which implies that $\|x-x^\perp\|\le\|x\|$.

Now $(3)$, Cauchy-Schwarz, and $(4)$ yield $$ \begin{align} \left|\sum_{i=1}^k\left<x,v_i\right>\left<y,v_i\right>\right| &=\left|\left<x-x^\perp,y-y^\perp\right>\right|\\ &\le\|x-x^\perp\|\|y-y^\perp\|\\ &\le\|x\|\|y\|\tag{5} \end{align} $$ To finish off the proof (thanks to cardinal), consider the vector $$ x^+=\sum_{i=1}^k\left|\left<x,v_i\right>\right|v_i\tag{6} $$ Note that $\|x^+\|^2=\sum\limits_{i=1}^k\left<x,v_i\right>^2=\|x-x^\perp\|^2$.

Plugging $x^+$ and $y^+$ into $(5)$ gives $$ \begin{align} \sum_{i=1}^k\left|\left<x,v_i\right>\left<y,v_i\right>\right| &=\left|\sum_{i=1}^k\left<x^+,v_i\right>\left<y^+,v_i\right>\right|\\ &\le\|x^+\|\|y^+\|\\ &\le\|x\|\|y\|\tag{7} \end{align} $$

$\endgroup$
12
  • $\begingroup$ ...and, to conclude, note that this holds for arbitrary $x$ and $y$, hence, it holds for $x^+ = \sum_{i=1}^k |\langle x, v_i \rangle | v_i$ and similarly defined $y^+$. Noting that $\|x^+\| \leq \|x\|$ and $\|y^+\| \leq \|y\|$ completes the proof. (+1) $\endgroup$
    – cardinal
    Nov 20 '11 at 18:10
  • $\begingroup$ @cardinal: isn't $(5)$ exactly what was asked? In fact, $x^+=x-x^\perp$ so I think your concerns are already handled. $\endgroup$
    – robjohn
    Nov 20 '11 at 18:13
  • $\begingroup$ The modulus is on the inside of the sum and $x^+ \neq x - x^\perp$ in general. $\endgroup$
    – cardinal
    Nov 20 '11 at 18:15
  • 1
    $\begingroup$ @LHS: Pythagoras. (Or, just check $\langle x^\perp, x-x^\perp \rangle = 0$ by the definitions.) $\endgroup$
    – cardinal
    Nov 20 '11 at 18:27
  • 1
    $\begingroup$ This is a very nice answer. $\endgroup$
    – cardinal
    Nov 20 '11 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.