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.
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#.
We apply this result to the case of #4# dimensions. Let #L:\mathbb{R}^4\to\mathbb{R}^4# be an orthogonal map which is not a scalar multiplication (so it is not equal to #\pm I_4#). Then there is an orthonormal basis #\alpha# of #\mathbb{R}^4# such that #L_\alpha# has one of the following forms where #\varphi# and #\psi# lie in the open interval #\ivoo{0}{\pi}#:
\[\begin{array}{rclcl}\text{two mutually perpendicular rotations}&:& L_\alpha &=&
\left(\,\begin{array}{cc}
D_{\varphi}&0\\
0&D_{\psi}
\end{array}\,\right)\\ \text{a single rotation}&:& L_\alpha &=&
\left(\,\begin{array}{cc}
I_2&0\\
0&D_{\varphi}
\end{array}\,\right)\\ \text{an improper rotation in 3D}&:& L_\alpha &=&
\left(\,\begin{array}{ccc}
1&0&0\\ 0&-1&0\\
0&0&D_{\varphi}
\end{array}\,\right)\\ \text{a point reflection in 2D perpendicular to a rotation}&:& L_\alpha &=&
\left(\,\begin{array}{cc}
-I_2&0\\
0&D_{\varphi}\end{array}\,\right)\\ \text{a point reflection in 2D}&:& L_\alpha &=&
\left(\,\begin{array}{cc}
I_2&0\\
0&-I_2
\end{array}\,\right)\\\text{a reflection}&:& L_\alpha &=&
\left(\,\begin{array}{cc}
I_3&0\\
0 &-1
\end{array}\,\right)\\\end{array}
\]
As before, we denote by #E_\lambda# the eigenspace of #L# with eigenvalue #\lambda#. We know that every eigenvalue has absolute value #1#.
Suppose that #\lambda# is a real eigenvalue of #L#. Then #\lambda=\pm1# and #E_{\lambda}^\perp# is an #L#-invariant subspace of #V# on which #L# has no eigenvalue #\lambda#. In particular, #W = E_{1}+E_{-1}# is an #L#-invariant subspace, so #W^\perp# is an #L#-invariant subspace of #V# with #V = W \oplus W^\perp#. The restriction of #L# to # W^\perp# has no real roots; rather it has #r\ge0# non-real roots with absolute value #1#, which we can write as\[\ee^{\ii\varphi_i}=\cos(\varphi_i)+\ii\sin(\varphi_i)\qquad\text{for }\varphi_i\in\ivoo{-\pi}{0}\cup\ivoo{0}{\pi}\text{ and }i=1,\ldots,r\] accompanied by an additional #r# non-real roots #\ee^{-\ii\varphi_1},\ldots,\ee^{-\ii\varphi_r}# with opposite imaginary parts. Since for each pair #\ee^{\ii\varphi_i},\ee^{-\ii\varphi_i}# of mutually complex conjugated roots it is arbitrary which one we call #\lambda_i# and which one we call #\overline{\lambda}_i#, we may choose to call a root #\lambda_i# if its imaginary part is positive and #\overline{\lambda}_i# if its imaginary part is negative. This means we may write #\lambda_i=\ee^{\ii\varphi_i}# in which the angle #\varphi_i# is restricted to the domain #\ivoo{0}{\pi}# such that #\sin(\varphi_i)# is positive for all #i=1,\ldots,r#.
First consider eigenvalue #\lambda_1# with non-real eigenvector #\vec{a}_1#. According to the theorem Real Jordan block, the map #L# has a #2#-dimensional #L#-invariant subspace #U_1# of #W^\perp# corresponding to #\lambda_1# such that the restriction of #L# to #U_1# has the matrix\[\matrix{\Re\lambda_1&-\Im\lambda_1\\\Im\lambda_1& \Re\lambda_1}=\matrix{\cos(\varphi_1)&-\sin(\varphi_1)\\\sin(\varphi_1)&\cos(\varphi_1)}=D_{\varphi_1}\] with respect to the orthonormal basis #\basis{\Re\vec{a}_1,-\Im\vec{a}_1}# for #U_1# for some real number #\varphi_1# in #\ivoo{0}{\pi}#. (With respect to the basis #\basis{\Re\vec{a}_1,\Im\vec{a}_1}# the matrix would be #D_{-\varphi_1}#.) The same reasoning with #W' = E_1+E_{-1}+U_1# instead of #W# shows that #(W')^\perp# has a #2#-dimensional #L#-invariant subspace #U_2# corresponding to eigenvalue #\lambda_2# and eigenvector #\vec{a}_2# such that the restriction of #L# to #U_2# has the matrix #D_{\varphi_2}# with respect to the orthonormal basis #\basis{\Re\vec{a}_2,-\Im\vec{a}_2}# for #U_2#, for a certain #\varphi_2# in #\ivoo{0}{\pi}#. Continuing this way, we find that #W^\perp# is the direct sum of #r# two-dimensional #L#-invariant subspaces \[W^\perp=U_1\oplus U_2\oplus\ldots\oplus U_r\]such that the restriction of #L# to each of these subspcaces is a rotation, that is to say: with respect to the orthonormal basis #\basis{\Re\vec{a}_i,-\Im\vec{a}_i}# for #U_i# the restriction of #L# to #U_i# has matrix #D_{\varphi_i}# for a real number #\varphi_i# in #\ivoo{0}{\pi}#.
Choose orthonormal bases for the eigenspaces #E_1 # and #E_{-1}#of the map #L#. Along with orthonormal bases for the #2#-dimensional #L#-invariant subspaces these bases provide an orthonormal basis for #V#. With respect to this basis, we find a matrix of #L# of the form
\[ J =
\left(\,\begin{array}{cccccccccc}
1\\
& \ddots\\
& & 1\\
& & & -1 & & & & 0 \\
& & & & \ddots\\
& & & & & -1\\
& & & & & & D_{\varphi_1}\\
& & & 0 & & & & D_{\varphi_2}\\
& & & & & & & & \ddots\\
& & & & & & & & & D_{\varphi_r}
\end{array}\,\right)
\] The uniqueness of the orthogonal Jordan normal form for a given #L# follows from the fact that each characteristic polynomial corresponds to exactly one orthogonal Jordan normal form, up to the order of the angles #\varphi_i#. Two Jordan normal forms with the same number of eigenvalues #1# and #-1# and the same angles (up to ordering) can be transformed into each other by means of a permutation of the vectors in the corresponding orthonormal basis; the matrix of such a permutation is a permutation matrix, and thus orthogonal.
A direct consequence of this form is the fact that orthogonal maps on finite-dimensional vector spaces are complex diagonalizable. After all, as we have seen in the comment on diagonalizability of Matrix of a #2#-dimensional orthogonal map, each submatrix # D_ {\varphi_j} # along the diagonal of # J # is complex diagonalizable, and so conjugate to the diagonal matrix with # \ee ^ {\varphi_j \ii} #, # \ee ^ {- \varphi_j \ii} # on the diagonal. If # T_j # is a conjugator, for instance #T_j =\matrix{1&\ii\\ 1&-\ii}#, it satisfies \[T_j\, D_{\varphi_j} \, T_j^{-1}=\matrix{\ee^{\varphi_j\ii}&0\\0&\ee^{-\varphi_j\ii}}\] Then conjugation of # J # by \[ T =
\left(\,\begin{array}{cccccccccc}
1\\
& \ddots\\
& & 1\\
& & & 1 & & & & 0 \\
& & & & \ddots\\
& & & & & 1\\
& & & & & & T_{1}\\
& & & 0 & & & & T_{2}\\
& & & & & & & & \ddots\\
& & & & & & & & & T_{r}
\end{array}\,\right)
\] gives the diagonal matrix \[ T\,J\,T^{-1} = \small
\left(\,\begin{array}{cccccccccc}
1\\
& \ddots\\
& & 1\\
& & & -1 & & & & 0 \\
& & & & \ddots\\
& & & & & -1\\
& & & & & & \matrix{\ee^{\varphi_1\ii}&0\\0&\ee^{-\varphi_1\ii}}\\
& & & 0 & & & &\matrix{\ee^{\varphi_2\ii}&0\\0&\ee^{-\varphi_2\ii}}\\
& & & & & & & & \ddots\\
& & & & & & & & &\matrix{\ee^{\varphi_r\ii}&0\\0&\ee^{-\varphi_r\ii}}
\end{array}\,\right)
\]
Each non-real eigenvalue of an orthogonal map is an eigenvalue of a rotation matrix #D_{\varphi_i}# and so has absolute value #1#.
Let #A # be an orthogonal matrix. According to the theorem, # A # is conjugate to a matrix as indicated above by means of a conjugator which is an orthogonal matrix. According to theorem Real Jordan normal form, this is the unique matrix (up to a permutation of the rotation matrices) of this form to which # A # is conjugate. This implies that, if # A # is conjugate to an orthogonal matrix # B #, there is an orthogonal conjugator; in other words, there is an orthogonal matrix # X #, such that # B = X \, A \, X ^ {- 1} #. Using the polar decomposition of matrices (which we will deal with later) we can give a more direct proof of this statement.
Consider the orthogonal #(4\times4)#-matrix \[ A = {{1}\over{50}}\,\matrix{-17 & -17 & 31 & -31 \\ 17 & -17 & 31 & 31 \\ 31 & -31 & -17 & -17 \\ 31 & 31 & 17 & -17 \\ }\] The linear mapping #L_A :\mathbb{R}^4\to\mathbb{R}^4#
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)#.
#\rv{\cos(\varphi) ,\cos(\psi)} =# #\rv{{{7}\over{25}},-{{24}\over{25}}}#
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:
\[\begin{array}{rcl}p_A(x)& =& \displaystyle x^4+{{34 x^3}\over{25}}+{{578 x^2}\over{625}}+{{34 x}\over{25}}+1\end{array}\] 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
\[\eqs{-2\cdot(a+b) &=&\displaystyle {{34}\over{25}}\\ 2+ 4a\cdot b &=& \displaystyle {{578}\over{625}}}\] A solution of this system is
\[\rv{a,b} =\rv{ {{7}\over{25}},-{{24}\over{25}}}\] The only other solution is obtained by interchanging the values of #a# and #b#, and we can omit it.
Jordan normal form for orthogonal maps
