### Orthogonal and symmetric maps: Classification of orthogonal maps

### Jordan normal form for orthogonal maps

After extension to the complex numbers, each orthogonal map on a finite dimensional inner product space is *diagonalizable*. This, and more, is proved in the following theorem.

Let #L# be an orthogonal map on a finite-dimensional inner product space #V#, and suppose that #W# is an #L#-invariant linear subspace of #V#. According to *property 5 of orthogonal maps*, the orthogonal complement of #W# is also invariant under #L#. Now choose an orthonormal basis #\alpha# for #V# consisting of an orthonormal basis of #W# complemented by an orthonormal basis of #W^\perp#. Then, the matrix of #L# with respect to #\alpha# has the form \[L_\alpha =\left(\,\begin{array}{cc}

M_1 & 0 \\ 0 & M_2

\end{array}\,\right)\] where #M_1# and # M_2# are orthogonal matrices of the maps #\left.L\right|_W:W\rightarrow W# and #\left.L\right|_{W^\perp} :W^\perp\rightarrow W^\perp#, respectively, and where #0# at each entry represents a zero matrix of appropriate dimensions.

We use this property to describe the *real Jordan normal form* of orthogonal matrices. It is not difficult to derive that a matrix in *real Jordan normal form* is orthogonal if and only if all its real eigenvalues are equal to #1# or #-1#, all #(2\times2)#-matrices along the diagonal are of the form \[D_{\varphi} = \matrix{\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi)& \cos(\varphi)} \] for certain #\varphi#, and no Jordan block has ones off the diagonal. The #(2\times2)#-matrices #D_{\varphi}# are the rotation matrices of *Orthogonal matrices in two dimensions*. The theorem below shows that every orthogonal matrix can be brought into this form by conjugation by an orthogonal matrix.

Orthogonal Jordan normal form Let #L:V\to V# be an orthogonal map on a finite-dimensional inner product space #V#. Then there is an orthonormal basis #\alpha# such that \[L_\alpha =\left(\,\begin{array}{cccccccccc}

1\\

& \ddots\\

& & 1\\

& & & -1 & & & & 0 \\

& & & & \ddots\\

& & & & & -1\\

& & & & & & D_{\varphi_1}\\

& & & 0 & & & & D_{\varphi_2}\\

& & & & & & & & \ddots\\

& & & & & & & & & D_{\varphi_r}

\end{array}\,\right)

\] where #\varphi_i\in\ivoo{0}{\pi}# is a real number for each #i# with #1\le i\le r#.

A matrix of this type is called an **orthogonal Jordan normal form**. The orthogonal Jordan normal form of #L# is uniquely determined up to the order of the matrices #D_{\varphi_i}#. We call #D_{\varphi_i}# for #i=1,\ldots,r# the **rotation matrices** of #L#.

*determined*by #A# consists of rotations in two planes passing through the origin which are perpendicular to each other. In one plane, there is a rotation over an angle #\varphi#; in the other plane a rotation over an angle #\psi#.

Determine the cosine of each of the two angles. Give your answer in the form of a list of two numbers #\rv{a,b}# where #a=\cos(\varphi)# and #b = \cos(\psi)#.

First we calculate the characteristic polynomial #p_A(x)# of #A#. This can be done by working out the formula #p_A(x) = \det(A-x\cdot I_4)#, or by calculating the powers of #A# and solving the linear equation, given by the

*Cayley-Hamilton theorem*,

\[A^4-\text{trace}(A)\cdot A^3+c\cdot A^2+d\cdot A+I_4 = 0\] with unknowns #c# and #d#. The result is:

\[\begin{array}{rcl}p_A(x)& =& \displaystyle x^4-{{10 x^3}\over{13}}+2 x^2-{{10 x}\over{13}}+1\end{array}\] Because #\mathbb{R}^4# is the direct sum of two planes passing through the origin which are invariant under #A#, the polynomial #p_A(x)# is the product of the characteristic polynomials of rotations in a plane over angles #\varphi# and #\psi#, so

\[\begin{array}{rcl} p_A(x)& =&(x^2-2\cos(\varphi)\cdot x +1) \cdot(x^2-2\cos(\psi)\cdot x +1 )\\

& =&x^4-2(a+b)\cdot x^3 +(2+4a\cdot b)\cdot x^2-2(a+b)\cdot x+1\end{array}\] where #a=\cos(\varphi)# and #b = \cos(\psi)#.

Comparison of these two expressions for #p_A(x)# tells us that

\[\eqs{-2\cdot(a+b) &=&\displaystyle -{{10}\over{13}}\\ 2+ 4a\cdot b &=& \displaystyle 2}\] A solution of this system is

\[\rv{a,b} =\rv{ 0,{{5}\over{13}}}\] The only other solution is obtained by interchanging the values of #a# and #b#, and we can omit it.

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