I'm trying to determine the ratio between $|\langle Av,v \rangle |$ and $\|v\|^2$ where $A$ is an $n\times n$ unitary matrix and $v\in \mathbb C^n$. In particular I'm trying to determine whether $|\langle Av,v \rangle | \le \|v\|^2$ or $|\langle Av,v \rangle | \gt \|v\|^2$.

This is what I have so far:

Let $w,u\in \mathbb C^n$ such that $w+u=v$ then:

Computing $|\langle Av,v \rangle |$ : $$|\langle Av,v \rangle | =|\langle A(w+u),w+u \rangle |=|\langle Aw+Au,w+u \rangle | = |\langle Aw,w \rangle + \langle Aw,u \rangle + \langle Au,w \rangle + \langle Au,u \rangle |$$

Computing $\|v\|^2$ :

$$\|v\|^2 = \langle v,v \rangle = \langle w+u,w+u \rangle = \langle w,w \rangle + \langle w,u \rangle + \langle u,w \rangle + \langle u,u \rangle$$

I'm not quite sure how to continue. I can use the exchange lemma and the fact that $A^{-1} = A^*$ because $A$ is an unitary matrix but I can't see how it helps me to simply the first equation.

Note : $\langle \cdot \rangle$ denotes the dot product.


3 Answers 3


Hint: A very famous theorem allows us to state (fill in the blanks)

$$ \left|\langle Av, v\rangle\right|\leqslant\left\|Av\right\|\left\|v\right\|\leqslant \left\|A\right\|\ldots\,, $$

And, don't forget that $A$ is unitary.

  • $\begingroup$ Just to make sure I understand correct : $$\left|\langle Av, v\rangle\right|\leqslant\left\|Av\right\|\left\|v\right\|\leqslant \left\|A\right\|\|v\|\|v\| = |A|\|v\|\|v\| $$ and $|A|=1$ because $A$ is unitary. Correct? $\endgroup$ Feb 9, 2015 at 10:11
  • $\begingroup$ @SyndicatorBBB, yep! :) $\endgroup$
    – ki3i
    Feb 9, 2015 at 10:12
  • $\begingroup$ First of all thank you very much! Very simple and nice solution and more important a basic theorem I should have known about. For some reason the fact that $|A|=1$ where $A$ is unitary wasn't native to me so I'll just make sure I understand it and that's it. Thank you again :) $\endgroup$ Feb 9, 2015 at 10:14
  • $\begingroup$ @SyndicatorBBB, just to be clear, $$\left\|A\right\|{}={}1$$ follows, in this case, since $A$ is unitary. Use the definition $$\left\|A\right\|=\sup_{|x|=1}\left\{\left\|Ax\right\|\right\}$$ $\endgroup$
    – ki3i
    Feb 9, 2015 at 10:43

$$|\langle Av,v\rangle|=|v^*A^Hv|\le_{CS} \|Av\|_2\|v\|_2=\|v\|_2^2,\ (\because\|Av\|_2^2=\|v\|_2^2\quad\mbox{by unitarity of }A)$$ CS is Cauchy-Scwartz inequality.

  • $\begingroup$ Why $\|Av\|_2\|v\|_2=\|v\|_2^2$? $\endgroup$ Feb 9, 2015 at 10:16
  • $\begingroup$ $\because \|Av\|_2=\|v\|_2$ $\endgroup$ Feb 9, 2015 at 10:18
  • $\begingroup$ I understand. Thank you very much :) $\endgroup$ Feb 9, 2015 at 10:20
  • $\begingroup$ Oh and one more thing isn't it $|\langle Av,v\rangle|=|v^TA^T\bar v|$? $\endgroup$ Feb 9, 2015 at 10:22
  • $\begingroup$ Oh, sorry, I wrote that with real vector in mind. I am editing it. $\endgroup$ Feb 9, 2015 at 10:32

You can do better than Cauchy Schwarz by observing that you are trying to maximize (i.e. find the maximum attained value) a quadratic form. Indeed, you are given the function $$ Q(v)=\langle Av, v\rangle$$ which is homogeneous of degree 2, that is $$ Q(zv)=\lvert z\rvert^2Q(v), \qquad \forall z\in \mathbb{C},\ \forall v\in \mathbb{C}^n,$$ and you are trying to find $$\tag{1}M=\max\left\{ |Q(v)|\, :\, v\in \mathbb{C}^n\ \lVert v\rVert =1\right\}.$$ (Homogeneity then gives the sharp inequality $|Q(v)|\le M\lVert v\rVert^2,\ v\in \mathbb{C}^n$). To solve the maximization problem (1) you can use the spectral theorem to unitarily diagonalize $A$: $$ A=U\,\mathrm{diag}(\lambda_1\ldots\lambda_n)U^\star, $$ where $U$ is unitary. Therefore $$ |Q(v)|=\left\lvert\sum_{j=1}^n \lambda_j \lvert w_j\rvert^2\right\rvert, \qquad w=U^\star v.$$ Since $U$ is unitary one has $\lVert w\rVert=\lVert v\rVert=1$. Moreover, $A$ is unitary too, which implies that all of its eigenvalues have modulus $1$: $\lvert\lambda_j\rvert=1$. We conclude that $$ |Q(v)|\le \sum_{j=1}^n \lvert\lambda_j\rvert \lvert w_j\rvert^2 = \sum_{j=1}^n \lvert w_j\rvert^2=1. $$ Therefore $M\le 1$. Evaluating $Q$ at $w=(1, 0, \ldots , 0)$ we see that actually $M=1$.

The advantage of this method over Cauchy Schwarz is that it not only gives a bound, it also shows that the bound is sharp. Moreover, the method is flexible and works for any kind of quadratic functions, even in infinite dimensions.


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