Since a unitary transformation on a unitary space is normal, the results of the preceding section include the theory of unitary transformations as a special case. Since, however, an orthogonal transformation on a real inner product space need not have any proper values, the spectral theorem, as we know it so far, gives us no information about orthogonal transformations. It is not difficult to get at the facts; the theory of complexification was made to order for this purpose.

Suppose that \(U\) is an orthogonal transformation on a finite-dimensional real inner product space \(\mathcal{V}\) ; let \(U^{+}\) be the extension of \(U\) to the complexification \(\mathcal{V}^{+}\) . Since \(U^{*} U=1\) (on \(\mathcal{V}\) ), it follows that \((U^{+})^{*} U^{+}=1\) (on \(\mathcal{V}^{+}\) ), that is, that \(U^{+}\) is unitary.

Let \(\lambda=\alpha+i \beta\) be a complex number ( \(\alpha\) and \(\beta\) real), and let \(\mathcal{M}\) be the subspace consisting of all solutions of \(U^{+} z=\lambda z\) in \(\mathcal{V}^{+}\) . (If \(\lambda\) is not a proper value of \(U^{+}\) , then \(\mathcal{M}=\mathcal{O}\) .) If \(z\) is in \(\mathcal{M}\) , write \(z=x+i y\) , with \(x\) and \(y\) in \(\mathcal{V}\) . The equation \[U x+i U y=(\alpha+i \beta)(x+i y)\] implies (cf. Section: Complexification ) that \[U x=\alpha x-\beta y\] and \[U y=\beta x+\alpha y.\] If we multiply the second of the last pair of equations by \(i\) and then subtract it from the first, we obtain \[U x-i U y=(\alpha-i \beta)(x-i y).\] This means that \(U^{+} \bar{z}=\bar{\lambda} \bar{z}\) , where the suggestive and convenient symbol \(\bar{z}\) denotes, of course, the vector \(x-i y\) . Since the argument (that is, the passage from \(U^{+} z=\lambda z\) to \(U^{+} \bar{z}=\bar{\lambda} \bar{z}\) ) is reversible, we have proved that the mapping \(z \to \bar{z}\) is a one-to-one correspondence between \(\mathcal{M}\) and the subspace \(\overline{\mathcal{M}}\) consisting of all solutions \(\bar{z}\) of \(U^{+} \bar{z}=\bar{\lambda} \bar{z}\) . The result implies, among other things, that the complex proper values of \(U^{+}\) come in pairs; if \(\lambda\) is one of them, then so is \(\bar{\lambda}\) . (This remark alone we could have obtained more quickly from the fact that the coefficients of the characteristic polynomial of \(U^{+}\) are real.)

We have not yet made use of the unitary character of \(U^{+}\) . One way we can make use of it is this. If \(\lambda\) is a complex (definitely not real) proper value of \(U^{+}\) , then \(\lambda \neq \bar{\lambda}\) ; it follows that if \(U^{+}{z}=\lambda z\) , so that \(U^{+} \bar{z}=\bar{\lambda} \bar{z}\) , then \(z\) and \(\bar{z}\) are orthogonal. This means that \[0=(x+i y, x-iy)=\|x\|^{2}-\|y\|^{2}+i\big((x, y)+(y, x)\big),\] and hence that \(\|x\|^{2}=\|y\|^{2}\) and \((x, y)=-(y, x)\) . Since a real inner product is symmetric ( \((x, y)=(y, x)\) ), it follows that \((x, y)=0\) . This, in turn, implies that \(\|z\|^{2}=\|x\|^{2}+\|y\|^{2}\) and hence that \(\|x\|=\|y\|=\frac{1}{\sqrt{2}}\|z\|\) .

If \(\lambda_{1}\) and \(\lambda_{2}\) are proper values of \(U^{+}\) with \(\lambda_{1} \neq \lambda_{2}\) and \(\lambda_{1} \neq \bar{\lambda}_{2}\) , and if \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=x_{2}+i y_{2}\) are corresponding proper vectors ( \(x_{1}\) , \(x_{2}\) , \(y_{1}\) , \(y_{2}\) in \(\mathcal{V}\) ), then \(z_{1}\) and \(z_{2}\) are orthogonal and (since \(\bar{z}_{2}\) is a proper vector belonging to the proper value \(\bar{\lambda}_{2}\) ) \(z_{1}\) and \(\bar{z}_{2}\) are also orthogonal. Using again the expression for the complex inner product on \(\mathcal{V}^{+}\) in terms of the real inner product on \(\mathcal{V}\) , we see that \[(x_{1}, x_{2})+(y_{1}, y_{2})=(x_{1}, y_{2})-(y_{1}, x_{2})=0\] and \[(x_{1}, x_{2})-(y_{1}, y_{2})=(x_{1}, y_{2})+(y_{1}, x_{2})=0.\] It follows that the four vectors \(x_{1}\) , \(x_{2}\) , \(y_{1}\) , and \(y_{2}\) are pairwise orthogonal.

The unitary transformation \(U^{+}\) could have real proper values too. Since, however, we know that the proper values of \(U^{+}\) have absolute value one, it follows that the only possible real proper values of \(U^{+}\) are \(+1\) and \(-1\) . If \(U^{+}(x+i y)= \pm(x+i y)\) , then \(U x= \pm x\) and \(U y= \pm y\) , so that the proper vectors of \(U^{+}\) with real proper values are obtained by putting together the proper vectors of \(U\) in an obvious manner.

We are now ready to take the final step. Given \(U\) , choose an orthonormal basis, say \(\mathcal{X}_{1}\) , in the linear manifold of solutions of \(U x=x\) (in \(\mathcal{V}\) ), and, similarly, choose an orthonormal basis, say \(\mathcal{X}_{-1}\) , in the linear manifold of solutions of \(U x=-x\) (in \(\mathcal{V}\) ). (The sets \(\mathcal{X}_{1}\) and \(\mathcal{X}_{-1}\) may be empty.) Next, for each conjugate pair of complex proper values \(\lambda\) and \(\bar{\lambda}\) of \(U^{+}\) , choose an orthonormal basis \(\{z_{1}, \ldots, z_{r}\}\) in the linear manifold of solutions of \(U^{+} z=\lambda z\) (in \(\mathcal{V}^{+}\) ). If \(z_{j}=x_{j}+i y_{j}\) (with \(x_{j}\) and \(y_{j}\) in \(\mathcal{V}\) ), let \(\mathcal{X}_{\lambda}\) be the set \(\{\sqrt{2} x_{1}, \sqrt{2} y_{1}, \ldots, \sqrt{2} x_{r}, \sqrt{2} y_{r}\}\) of vectors in \(\mathcal{V}\) . The results we have obtained imply that if we form the union of all the sets \(\mathcal{X}_{1}\) , \(\mathcal{X}_{-1}\) , and \(\mathcal{X}_{\lambda}\) , for all proper values \(\lambda\) of \(U^{+}\) , we obtain an orthonormal basis of \(\mathcal{V}\) . In case \(\mathcal{X}_{1}\) has three elements, \(\mathcal{X}_{-1}\) has four elements, and there are two conjugate pairs \(\{\lambda_{1}, \bar{\lambda}_{1}\}\) and \(\{\lambda_{2}, \bar{\lambda}_{2}\}\) , then the matrix of \(U\) with respect to the basis so constructed looks like this: \begin{align} \setcounter{MaxMatrixCols}{15} \begin{bmatrix} 1 & & & & & & & & & & \\ & 1 & & & & & & & & & \\ & & 1 & & & & & & & & \\ & & & -1 & & & & & & & \\ & & & & -1 & & & & & & \\ & & & & & -1 & & & & & \\ & & & & & & -1 & & & & \\ & & & & & & & \alpha_1 & -\beta_1 & & \\ & & & & & & & \beta_1 & \alpha_1 & & \\ & & & & & & & & & \alpha_2 & -\beta_2\\ & & & & & & & & & \beta_2 & \alpha_2 \end{bmatrix} \end{align} (All terms not explicitly indicated are equal to zero.) In general, there is a string of \(+1\) ’s on the main diagonal, followed by a string of \(-1\) ’s, and then there is a string of two-by-two boxes running down the diagonal, each box having the form \(\big[\begin{smallmatrix} \alpha & -\beta \\ \beta & \alpha \end{smallmatrix}\big]\) , with \(\alpha^{2}+\beta^{2}=1\) . The fact that \(\alpha^{2}+\beta^{2}=1\) implies that we can find a real number \(\theta\) such that \(\alpha=\cos \theta\) and \(\beta=\sin \theta\) ; it is customary to use this trigonometric representation in writing the canonical form of the matrix of an orthogonal transformation.

EXERCISES