MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x, y\in R^n$ and $x, y$ are nonzero, it is well known $\frac{x^Ty}{\parallel x\parallel_2\parallel y\parallel_2}(\parallel x\parallel_2+\parallel y\parallel_2)\le \parallel x+y\parallel_2$. How to extend this to complex vectors? $arcos\frac{x^Ty}{\parallel x\parallel_2\parallel y\parallel_2} $ is the angle between $x$ and $y$. What is the appropriate definition of the angle between two complex vectors? I know $\frac{\mid x^*y\mid}{\parallel x\parallel_2\parallel y\parallel_2}(\parallel x\parallel_2+\parallel y\parallel_2)\le \parallel x+y\parallel_2$ does not hold generally.

share|cite|improve this question
up vote 3 down vote accepted

Let $x,y$ be two nonzero complex vectors, let $\hat x=x/\|x\|$ and $\hat y=y/\|y\|$, and consider the parabola $$\phi(t)=\|t\hat x+(1-t)\hat y\|^2=1+2(t^2-t)(1-\Re(\hat x \overline{\hat y})). $$ You easily check that $\phi(t)\ge\phi(1/2)$ for all $t$. This gives the inequality $$ \|t\hat x+(1-t)\hat y\|\ge \sqrt{\frac{1+\Re(\hat x \overline{\hat y})}2} $$ for all real $t$. This is stronger than your inequality, which can be obtained by choosing $$t=\frac{\|x\|}{\|x\|+\|y\|}$$ at the left hand side, and noticing that $$ \sqrt{\frac{1+\sigma}2}\ge\sigma $$ for all real $|\sigma|\le1$ at the right hand side. So yes, the correct extension is using $\Re(x \cdot\overline{y})$ instead of $x\cdot y$.

share|cite|improve this answer
This is just the cosine of the angle between the two vectors as real vectors. There is a more complex version of the angle between to complex vectors. The correspond to points in $\mathbb{C}P(n-1)$ and span a copy of $\mathbb{C}P(1)$. The copy of $\mathbb{C}P(1)$ is a round sphere of radius $1/2$ in the Fubini study metric. The complex angle between the two vectors is the arc length of the shortest geodesic on $\mathbb{C}P(1)$ joining them. The complex angle takes on values between $0$ and $\pi/2$. – Charlie Frohman Oct 12 '10 at 13:36
My point was to find the complex substitute in the inequality proposed by miwa – Piero D'Ancona Oct 12 '10 at 16:46

Please refer

K. Scharnhorst, “Angles in complex vector spaces,” Acta Applicandae Math., vol. 69, pp. 95–103, Nov. 2001.

and appendix in

V. G. Reju, S. N. Koh and I. Y. Soon, “Underdetermined Convolutive Blind Source Separation via Time-Frequency Masking,” IEEE Transactions on Audio, Speech and Language Processing, Vol. 18, NO. 1, Jan. 2010, pp. 101–116.

share|cite|improve this answer
Dear Reju thank you for that Scharnhorst reference - it has saved me an immense amount of time-wasting in my research! I'm surprised that stuff is not better known ... – user28127 Nov 15 '12 at 13:26

Actually, the case of complex vector spaces is rather a particular case than an extension, with respect to the case of real vector spaces. Recall that, as a vector space over $\mathbb{R}$, your $\mathbb{C}^n$ is isomorphic to $\mathbb{R}^{2n}$, and that, in terms of the Hermitian form of the former, the standard scalar product of the latter writes $\Re(x\cdot \bar y)$.

Generally speaking, the appropriate definition of topological/uniform/metric notions for complex vector spaces is just the same for real vector spaces, seeing the complex vector spaces as real vector spaces by restriction of scalars. So the angle of vectors in $\mathbb{C}^n$ is just the angle in $\mathbb{R}^{2n}$. Sometimes, in the complex version, one also requires some kind of algebraic compatibility with the complex structure (e.g., the definition of a norm for complex VS).

share|cite|improve this answer

Let $V$ be a Euclidean vector space (in particular $V$ can be a Hermitean vector space considered as a real vector space). According to J.H.C. Whitehead (Manifolds with transversal fields in Euclidean space, Ann Math 73, 154-212) the angle between vector subspaces $V_1$ and $V_2$ of $V$ can be defined as the Hausdorff distance (see e.g. between their intersections with the unit sphere. More explicitly, it is $$max(max_{x\in V_1\cap S_0}\angle(x,V_2),max_{x\in V_2\cap S_0}\angle(x,V_1))$$ where $S_0$ is the unit sphere and $\angle(x,W)$ is the angle between $x$ and its orthogonal projection to $W$ (if $x$ is orthogonal to $W$, the angle is set to be $\frac{\pi}{2}$).

share|cite|improve this answer
It is not easy to determine the angle in terms of entries of $x$ and $y$. – Sunni Oct 1 '10 at 2:42

Probably use inner product Re(x* y), where x* is the conjugate transpose of x ...

share|cite|improve this answer
Note that with this setup, the "angle" operator is no longer commutative; take care with using this. – J. M. Oct 1 '10 at 0:33
@ J.M, it is still commutative, since $Re(x^* y)=Re(y^* x)$. But I don't know whether $\frac{Re x^*y }{\parallel x\parallel_2\parallel y\parallel_2}(\parallel x\parallel_2+\parallel y\parallel_2)\le \parallel x+y\parallel_2$ is true. – Sunni Oct 1 '10 at 1:06
J.M.: it may not look commutative, but it is. consider the symmetrized version of the inner product <x,y> = $\frac{1}{2}$(x*y + xy*) = $\frac{1}{2}$(x*y + [x*y]*) = Re(x*y). this inner-product is just the usual inner-product on $\mathbb R ^2$, treating x and y as vectors in $\mathbb R ^2$: x $\to$ (Re(x), Im(x)). so the inequality for $\mathbb R ^2$ continues to hold for $\mathbb C$. – ronaf Oct 1 '10 at 1:28
I missed the $\Re$ on first reading; thanks for pointing it out. :) – J. M. Oct 1 '10 at 3:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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