# Help identifying a theorem about representations of an orthogonal linear transformation

I'm looking for some information on a theorem presented in my linear algebra notes stated without name - basically it states that (taken verbatim):

If $f: V \rightarrow V$ is an orthogonal linear transformation on a finite dimensional real inner product space, there exists an orthonormal basis of V of the form:

$$(u_1,v_1,u_2,v_2,...,u_k,v_k,w_1,...,w_l)$$

(I am slightly confused as to why the basis was presented in this form)

And it goes on to state that the matrix of the transformation can be represented as being 'block diagonal', with blocks of the form:

$$\begin{bmatrix} cos(\theta_k) & -sin(\theta_k) \\ sin(\theta_k) & cos(\theta_k) \\ \end{bmatrix}$$

or $\pm1$ along the diagonal.

There is no proof or name given and I was wondering if someone could recognize this theorem so I could find an alternative resource that gives a detailed explanation and proof of this theorem.

This is the existence of the canonical form of an orthogonal matrix. The proof boils down to showing that all of the eigenvalues of an orthogonal matrix have modulus $1$: Its real eigenvalues are $\pm 1$, and lead to the diagonal elements $\pm 1$. Since (real) orthogonal matrices are real, the nonreal eigenvalues come in complex conjugate pairs $\cos \theta_j \pm i \sin \theta_j$, and block diagonalizing w.r.t. the sums of eigenspaces for such pairs gives blocks of the form $$\begin{pmatrix}\cos \theta_j & -\sin \theta_j\\ \sin \theta_j & \cos \theta_j\end{pmatrix}.$$

• By block diagonalizing, do you just mean to find a basis to which the matrix of the linear transformation $f$ in two dimensions is of that form? – Eweler Oct 13 '14 at 12:02
• I wasn't speaking very precisely, but yes, one takes $1$-d eigenspaces $E_{\pm}$ corresponding to a pair of conjugate eigenvalues $\cos \theta_j \pm i \sin \theta_j$, and in an orthogonal basis $(u_j, v_j)$ of $E_j = E_j^+ \oplus E_j^-$, the restriction of the orthogonal matrix to $E_j$ is given by the above $2\times 2$ matrix (or the matrix replacing $\theta_j$ with $-\theta_j$, depending on some choices). In particular, the above block appears in the canonical form corresponding to the basis elements $(u_j, v_j)$ in the guaranteed adapted basis. – Travis Oct 13 '14 at 12:12

If you choose an orthonormal basis of $V$ and view $f$ as a matrix it is an orthogonal matrix. In the wikipedia article you'll see that the form you mention is called the canonical form. There is a stackexchange question but I'm not sure how helpful it is. In any case, I hope knowing the terminology will make it easy to find a proof.

• Thanks, I'll start reading up on it. – Eweler Oct 13 '14 at 9:25