# Special orthogonal matrices have orthogonal square roots

Let $A$ be an orthogonal matrix with $\det (A)=1$. Show that there exists an orthogonal matrix $B$ such that $B^2=A$.

Thank you very much.

Edit: This is indeed true...Thanks @Dirk.

An orthogonal matrix (see the "Canonical form" paragraph or this thread exhibited by user1551) $A$ is block diagonalizable in an orthonormal basis with blocks $$\left(\matrix {\cos\theta&-\sin\theta\\\sin\theta&\cos\theta}\right)$$ or $\pm 1$ along the diagonal, i.e. $P^*AP=A'$ with $P$ orthogonal and $A'$ block diagonal of rotations as above and $\pm 1$. I denote $M^*$ the adjoint matrix of $M$, which is nothing but the transpose $M^T$ in the real case.

If $\det A=1$, there are 2$k$ numbers $-1$ on the diagonal, corresponding to $k$ rotations of angle $\pi$. So we only have rotations and $1$'s.

Now, as pointed by @Dirk in the $-1$ case,

$$\left(\matrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta} \right)=\left(\matrix{\cos\theta/2&-\sin\theta/2\\\sin\theta/2&\cos\theta/2} \right)^2.$$

The genuine idea is that a rotation, is the square of the half-angle rotation. This works for $-I_2$ in particular, with $\theta=\pi$.

Do that as many times as needed to get $A'=C^2$ with $C$ orthogonal. Then define $$B:=PCP^*\qquad\Rightarrow\qquad B^2=PCP^*PCP^*=PC^2P^*=PA'P^*=A$$ and note that $B^*B=I$, so $B$ is orthogonal.

• I fixed my solution too, but you show it is unitary, but we need orthogonal don't we ? – Dominic Michaelis Apr 7 '13 at 12:02
• @DominicMichaelis No, when $B$ is real $B^*=B^T$. Orthogonal=real unitary=isometry. – Julien Apr 7 '13 at 12:03
• @XLDD Some key points about orthogonal matrices $A$ in $M_n(\mathbb{R})$. First, this means $A$ invertible with $A^{-1}=A^T$ the transpose. Second, a fortiori, as $\overline{A}=A$ is real, this implies $A^{-1}=A^*$ the adjoint, i.e. $A$ is unitary as an element of $M_n(\mathbb{C})$. So $A$ is diagonalizable via a unitary change of basis in $M_n(\mathbb{C})$, like every normal matrix. Third, the spectrum of $A$ is contained in the unit circle $\{z\in\mathbb{C};\;;|z|=1\}$. – Julien Apr 8 '13 at 10:26
• @XLDD Now since the characteristic polynomial $\chi_A$ of $A$ is real, this gives real eigenvalues $\pm 1$ with any possible multiplicity (corresponding to the linear factors of $\chi_A$), and complex eigenvalues $e^{i\theta}$ appearing as many times as their conjugate $e^{-i\theta}$ in the eigenvalue list (corresponding to the quadratic factors of $\chi_A$). The eigenvalues $\pm 1$ yield the real-diagonalizable part of $A$: they give rise to a diagonal of $\pm 1$ in an orthonormal basis. The complex ones correspond to the non-real-diagonalizable part. They give rise to $2\times 2$ blocks. – Julien Apr 8 '13 at 10:35
• @XLDD These are $\theta$ blocks of the form I considered in my answer above. THese are rotations of angle $\theta$. Note that, of course, a nontrivial rotation, say with angle $\pi/4$, can't leave any real line invariant: so these blocks are not real-diagonalizable. Finally, the major difficulty of this exercise is to show the statement for such blocks, as it is trivial for the diagonal elements $1$, and as the $-1$'s are in even number in this case (since $\det A=1$) and give rise to $\pi$ rotations which can be dealt with simultaneously. – Julien Apr 8 '13 at 10:41

Edit: I missunterstood the question at first thanks @Dirk again.

Thats true, $A$ is diagonalizable and with $\pm 1$ on the diagonal. you get an even number of $-1$ because else the determinant can't $1$. For this the matrix $B$ will be something like $$B= \begin{pmatrix} I_n &0 &0\\ 0 & 0 &1\\ 0& -1 & 0 \\ \end{pmatrix}$$ or more of the $$\begin{pmatrix} 0 & 1\\ -1 &0 \\\end{pmatrix}$$ blocks.

So we have $$A=P^{-1} D P = P^{-1} B B P= (P^{-1} B P) (P^{-1} B P)$$ And $B^T B=I$ hence $B$ is orthogonal.

• argh maybe we should stay in contact which questions we answer ;) – Dominic Michaelis Apr 7 '13 at 11:30
• I propose I only answer even numbered users. You get the odds. – Julien Apr 7 '13 at 11:31
• No, I meant the displayed registration number of the user. But this does not apply to those with nicknames. For these, I get the ones with an even number of characters. Easy. – Julien Apr 7 '13 at 11:35
• DominicMichaelis, @julien: with $B=[0\ 1;-1\ 0]$ it holds that $B^2 = -I$. – Dirk Apr 7 '13 at 11:42
• every normal operator is diagonalizable and $A$ is normal because it is orthogonal – Dominic Michaelis Apr 7 '13 at 12:13

Depending on what you know, the proof varies from a one-liner to a half page.

A one-line proof is this: every real orthogonal matrix with determinant $1$ can be written as $e^K$ for some skew symmetric matrix $K$; therefore $A=B^2$ for $B=e^{K/2}$.

A longer proof is this: $A$ is real and normal. Hence it is orthogonally similar to its real Jordan form. That is, $$A=Q\left(I_p\oplus-I_q\oplus R(\theta_1)\oplus\cdots\oplus R(\theta_m)\right)Q^T$$ where $Q$ is orthogonal, $\theta_1,\ldots,\theta_m\in(0,\pi)$, $R(\theta)$ denotes the $2\times2$ rotation matrix for angle $\theta$, and $p+q+2m=n$. (See this related question for the reason why such a decomposition is possible.) Since $\det A=1$, $q$ must be an even number. Therefore $A=B^2$, where $$B=Q\left(I_p\oplus \underbrace{R(\frac{\pi}2)\oplus \cdots R(\frac{\pi}2)}_{\frac q2 \text{ blocks}} \oplus R(\frac{\theta_1}2)\oplus\cdots\oplus R(\frac{\theta_m}2)\right)Q^T.$$

• Nice compact way of writing this down, +1. – Julien Apr 8 '13 at 13:49