Some properties of orthogonal maps

Orthogonal and symmetric maps: Orthogonal maps

Some properties of orthogonal maps

Here are some general properties of orthogonal maps.

Properties of orthogonal maps Let #V# be a real inner product space and #L:V\to V# an orthogonal map.

If also #M :V\rightarrow V# is an orthogonal map, then the composition #L\,M# is orthogonal too.
The map #L# is injective.
If #V# is finite dimensional, then #L# is invertible and #L^{-1}# is orthogonal.
Each real eigenvalue of #L# equals #1# or #-1#.
If #W# is a finite-dimensional linear subspace of #V# which is invariant under #L#, then also the orthogonal complement #W^\perp# is invariant under #L#.
If #L# fixes the orthogonal complement of a nonzero vector #\vec{v}# of #V#, then #L# is either the identity or the orthogonal reflection #S_{\vec{v}}#.

1. Because #L# and #M# are orthogonal, we have, for all #\vec{x},\vec{y}\in V#, \[\norm{L\,M(\vec{x})-L\,M(\vec{y} )}=\norm{M(\vec{x}) -M(\vec{x})}=\norm{\vec{x}- \vec{y}}\] This shows that #L\,M# is orthogonal.

2. Suppose #L(\vec{x}) = L(\vec{y}) #. Then we have \( L(\vec{x}) - L(\vec{y})=\vec{0}\), so #\norm{L(\vec{x}) - L(\vec{y})}=0#. Because #L# is orthogonal, it follows that #\norm{\vec{x} - \vec{y}}=0#, and, because of the positive-definiteness of the norm, #\vec{x} -\vec{y}=\vec{0}# so \( \vec{x}=\vec{y} \). This proves that #L# is injective.

3. First we show that #L# is invertible. The map #L# is finite-dimensional and linear, and, according to statement 2, also injective. From Invertibility criteria for a linear map it follows that #L# is invertible. All #\vec{x}\in V# satisfy #\vec{x}=L\, L^{-1}(\vec{x})#. Because #L# is orthogonal, it follows that \[ \norm{\vec{x}}=\norm{L\,L^{-1}(\vec{x})}=\norm{L^{-1}(\vec{x})}\] so #L^{-1}# is orthogonal.

4. Suppose that #\lambda# is a real eigenvalue of an orthogonal map #L:V\rightarrow V# and let #\vec{v}# be an eigenvector of #L# corresponding to #\lambda#. Then \[\abs{\lambda}\cdot\norm{\vec{v}}=\norm{\lambda\vec{v}}=\norm{L(\vec{v})}=\norm{\vec{v}}\] Because #\vec{v}\ne\vec{0}#, positivity of the norm gives that #\norm{\vec{v}} \gt0#, so #\abs{\lambda}=1#. It follows that #\lambda =1# or #\lambda = - 1#.

5. Suppose that #W# is a finite-dimensional #L#-invariant linear subspace of #V#. Then the restricition of #L# to #W# is an invertible map #W\to W# by statement 3. Let #\vec{x}# be a vector of #W^\perp#. Then for each vector #\vec{w}# of #W#\[\begin{array}{rcl}\dotprod{\vec{w}}{L(\vec{x})} &=& \dotprod{(L\,L^{-1}(\vec{w}))}{L(\vec{x})}\\ &&\phantom{xx}\color{blue}{\left.L\right|_W\text{ is invertible}}\\ &=& \dotprod{(L^{-1}(\vec{w}))}{\vec{x}}\\ &&\phantom{xx}\color{blue}{L\text{ is orthogonal}}\\ &=& 0\\ &&\phantom{xx}\color{blue}{L^{-1}(\vec{w})\text{ belongs to }W\text{ and }\vec{x}\text{ to }W^\perp}\end{array}\] This shows that \(L(\vec{x})\) is perpendicular to each vector #\vec{w}# of #W#, that is, \(L(\vec{x})\) belongs to #W^\perp#.

6. If #L# satisfies the hypotheses, then #\vec{v}# is an eigenvector of #L#. By 4, the corresponding eigenvalue is #1# or #-1#. In the former case, #L# is the identity. In the latter case, #L # is the unique linear map sending #\vec{v}# to #-\vec{v}# and fixing each vector in the orthogonal complement #\{\vec{v}\}^{\perp}#. This means that #L = S_{\vec{v}}#.

Infinite dimension In infinite-dimensional inner product spaces there exist orthogonal maps that have no inverse. Here is an example. Let #P# be the vector space of all polynomials in #x#. This is an inner product space with respect to the inner product for which #\basis{1,x,x^2,\ldots}# is orthonormal. Let #L:P\to P# be the map consisting of multiplication by #x#. Then #L# is orthogonal and injective, but not bijective.

Complex eigenvalues As we will see later, all complex eigenvalues of an orthogonal map have absolute value #1#.

Determinant Since the determinant is the product of all eigenvalues (with the appropriate multiplicities) and since non-real eigenvalues occur in pairs of mutually complex conjugate eigenvalues with absolute value #1#, it follows from statement 4 of the theorem that the determinant of an orthogonal map on a finite-dimensional inner product space equals #1# or #-1#. We will come back to this later.

Counterexample By means of a counterexample we show that the requirement in statement 5 that the subspace #W# be finite-dimensional, is necessary. Let #P# be the inner product space of all polynomials in #x# with #\basis{1,x,x^2,\ldots}# as orthogonal basis. The map #L_x:P\to P# which consists of multiplication by #x#, is clearly orthogonal. The linear subspace #W# consisting of all polynomials without constant term is clearly an infinite-dimensional #L_x#-invariant subspace of #P# with orthogonal complement #W^\perp = \linspan{1}#. But #L_x(1) = x# does not lie in #W^\perp#, so #W^\perp# is not #L_x#-invariant.

Suppose that #L: \mathbb{R}^3\to \mathbb{R}^3# is an orthogonal map with

\(L(\rv{1,1,-1}) = \rv{1,1,-1}\)
\(\rv{-1,0,-1}\) is an eigenvector with eigenvalue #-1#
The dimension of the eigenspace of #L# corresponding to eigenvalue #1# equals #1#.

Determine the matrix #A# of #L#.

#A = # # \matrix{-{{1}\over{3}} & {{2}\over{3}} & -{{2}\over{3}} \\ {{2}\over{3}} & -{{1}\over{3}} & -{{2}\over{3}} \\ -{{2}\over{3}} & -{{2}\over{3}} & -{{1}\over{3}} \\ }#

The vector #\rv{1,1,-1}# is fixed by #L# and so lies in the eigenspace with respect to the eigenvalue #1#. Because #L# is orthogonal, the only real eigenvalues are #1# and #-1#. The eigenspace with respect to the eigenvalue #-1# is either #1#-dimensional or #2#-dimensional. In the first case there would be a non-real eigenvalue, and so its complex conjugate would also be an eigenvalue, which is impossible because the dimension of the inner product space is equal to #3#. Therefore, the eigenspace of #L# with respect to #-1# has dimension #2#.

We find a vector perpendicular to both \( \rv{1,1,-1}\) and \(\rv{-1,0,-1}\). This can be found by solving a set of linear equations. A faster method uses the cross product:
\[\rv{1,1,-1}\times \rv{-1,0,-1} = \rv{-1,2,1 }\] This should be an eigenvector of #L# with eigenvalue #-1#. Thus we find that the matrix #L_\beta# of #L# relative to the basis
\[\beta = \basis{\rv{1,1,-1},\rv{-1,0,1},\rv{-1,2,1 }}\] is the diagonal matrix with diagonal entries #1#, #-1#, #-1#. We conclude that the matrix of #L# (relative to the standard basis #\varepsilon#) is equal to
\[\begin{array}{rcl} L_{\varepsilon} &=& {}_\varepsilon I_\beta \,L_\beta \, {}_\beta I_\varepsilon\\
&=&\matrix{1 & -1 & -1 \\ 1 & 0 & 2 \\ -1 & -1 & 1 \\ }\,\matrix{1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ } \,\matrix{1 & -1 & -1 \\ 1 & 0 & 2 \\ -1 & -1 & 1 \\ }^{-1} \\
&=&\matrix{1 & 1 & 1 \\ 1 & 0 & -2 \\ -1 & 1 & -1 \\ }\, \matrix{{{1}\over{3}} & {{1}\over{3}} & -{{1}\over{3}} \\ -{{1}\over{2}} & 0 & -{{1}\over{2}} \\ -{{1}\over{6}} & {{1}\over{3}} & {{1}\over{6}} \\ } \\
&=& \matrix{-{{1}\over{3}} & {{2}\over{3}} & -{{2}\over{3}} \\ {{2}\over{3}} & -{{1}\over{3}} & -{{2}\over{3}} \\ -{{2}\over{3}} & -{{2}\over{3}} & -{{1}\over{3}} \\ }
\end{array}\]

New example