Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Is the set of all unitary matrices in $M_2(\mathbb{C})$ is compact? I can show that as determinant map is continuous so unitary matrices are closed but how to show they are bounded?

Please help.

share|cite|improve this question
up vote 9 down vote accepted
  • $M_n(\mathbb C)$ is a finite-dimensional space, so it's enough to show that $U_n(\mathbb C)$ is closed and bounded.
  • The maps $f_1\colon U\mapsto U^*U-I$ and $f_2\colon U\mapsto UU^*-I$ are continuous since so is the map $U\mapsto U^*$, so $U_n(\mathbb C)=f_1^{-1}(\{0_n\})\cap f_2^{-1}(\{0_n\})$ is closed as an intersection of such two sets.
  • $U_n(\mathbb C)$ is bounded for the euclidian operator norm, since for each $x$ and $U$ unitary $$\langle x,x\rangle=\langle U^*Ux,x\rangle=\langle Ux,Ux\rangle$$ (hence $U$ is an isometry, in particular its norm is $1$).
share|cite|improve this answer

They are isometries for the hermitian form (i.e. $u^* M^*Mv=u^*v,\ \forall u,v\in\mathbb{C}^n$), so their operatorial norm is $1,$ (i.e. $||M||:=\sup_{|u|=1}|Mu|=1$.)
Hence $U(n)$ is included in the unit sphere of the normed vector space $(\mathfrak{gl}(\mathbb{C},n),||\cdot||).$

share|cite|improve this answer

One of the definitions of a unitary matrix is that its rows (or columns) form an orthonormal basis with respect to the standard inner product on $\mathbb{C}^n$; the set of orthonormal frames in $\mathbb{C}^n$ is obviously bounded. Topologically it's a torus, I think. [?]

This makes sense to me, since the eigenvalues of unitary matrices lie on the unit circle.

share|cite|improve this answer
Sorry to be late to the party. Topologically, $U(2)$ is $S^3\times S^1$, so it's not a torus in the usual sense. – Jason DeVito May 15 '12 at 0:39

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.