Some properties of orthogonal maps

Orthogonal and symmetric maps: Orthogonal maps

Some properties of orthogonal maps

Here are some general 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 $L = S_{\vec{v}}$ .

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.

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

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.

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

$L(\rv{2,3,2}) = \rv{2,3,2}$
$\rv{2,0,-2}$ is an eigenvector with eigenvalue $-1$
The dimension of the eigenspace of $L$ corresponding to eigenvalue $1$ equals $1$ .

Determine the matrix $A$ of $L$ .

$A =$ $\matrix{-{{9}\over{17}} & {{12}\over{17}} & {{8}\over{17}} \\ {{12}\over{17}} & {{1}\over{17}} & {{12}\over{17}} \\ {{8}\over{17}} & {{12}\over{17}} & -{{9}\over{17}} \\ }$

The vector $\rv{2,3,2}$ 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 $\rv{2,3,2}$ and $\rv{2,0,-2}$ . This can be found by solving a set of linear equations. A faster method uses the cross product:

$\rv{2,3,2}\times \rv{2,0,-2} = \rv{-6,8,-6 }$ This should be an eigenvector of $L$ with eigenvalue $-1$ . Thus we find that the matrix $L_\beta$ of $L$ relative to the basis

$\beta = \basis{\rv{2,3,2},\rv{2,0,2},\rv{-6,8,-6 }}$ 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

$\begin{array}{rcl} L_{\varepsilon} &=& {}_\varepsilon I_\beta \,L_\beta \, {}_\beta I_\varepsilon\\ &=&\matrix{2 & 2 & -6 \\ 3 & 0 & 8 \\ 2 & -2 & -6 \\ }\,\matrix{1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ } \,\matrix{2 & 2 & -6 \\ 3 & 0 & 8 \\ 2 & -2 & -6 \\ }^{-1} \\ &=&\matrix{2 & -2 & 6 \\ 3 & 0 & -8 \\ 2 & 2 & 6 \\ }\, \matrix{{{2}\over{17}} & {{3}\over{17}} & {{2}\over{17}} \\ {{1}\over{4}} & 0 & -{{1}\over{4}} \\ -{{3}\over{68}} & {{1}\over{17}} & -{{3}\over{68}} \\ } \\ &=& \matrix{-{{9}\over{17}} & {{12}\over{17}} & {{8}\over{17}} \\ {{12}\over{17}} & {{1}\over{17}} & {{12}\over{17}} \\ {{8}\over{17}} & {{12}\over{17}} & -{{9}\over{17}} \\ } \end{array}$

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