8. Isometries and self-adjoint mappings

Part 1.: Problems, solutions.
Part 2.: Problems, solutions.
Part 3.: Problems, solutions.

Isometries of linear euclidean spaces

A linear mapping \varphi\colon V_1\to V_2, where \langle V_1, \langle\cdot,\cdot \rangle_1\rangle, \langle V_2, \langle\cdot,\cdot \rangle_2\rangle are linear euclidean spaces is an isometry if the following equivalent conditions hold

  • is a linear isomorphism and preserves the inner product

        \[\forall_{v,w\in V} \langle v,w\rangle_1=\langle \varphi(v),\varphi(w)\rangle_2,\]

  • is a linear isomorphism and preserves the lengths of vectors

        \[\forall_{v\in V} \|v\|_1=\|\varphi(v)\|_2,\]

  • takes an orthonormal basis of V_1 onto a orthonormal basis of V_2,
  • takes any orthonormal basis of V_1 onto a orthonormal basis of V_2,

Orthogonal matrices

A matrix A is orthogonal if A\cdot A^T=I.

In other words, a matrix is orthogonal if its rows (equivalently columns) form an orthonormal basis with respect to the standard inner product.

Orthogonal matrices are invertible and A^T=A^{-1} for such matrices.

Notice that \varphi\colon V_1\to V_2 is an isometry if and only if M(\varphi)_{\mathcal{A}}^{\mathcal{B}} is orthogonal for some (equivalently, any) orthogonal bases \mathcal{A}, \mathcal{B} of respectively spaces V_1 and V_2.

Rotations and perpendicular reflection

If \varphi\colon V\to V is a perpendicular reflection across W, and if \mathcal{A} consists of an orthonormal basis of W after which we have an orthonormal basis of W^\bot we get

    \[M(\varphi)_{\mathcal{A}}^{\mathcal{A}}=\left[\begin{array}{cccccccc}1&0&\ldots &0&0&0&\ldots&0\\0&1&\ldots &0&0&0&\ldots&0\\&\ldots &&\ldots &&\ldots&&\ldots \\0&0&\ldots&1&0&0&\ldots&0\\0&0&\ldots&0&-1&0&\ldots&0\\ 0&0&\ldots&0&0&-1&\ldots&0\\&\ldots &&\ldots &&\ldots&&\ldots \\0&0&\ldots&0&0&0&\ldots&-1\end{array}\right].\]

Obviously a perpendicular reflection is an isometry.

If W is a two-dimensional subspace of V, then the rotation around W^{\bot} by \alpha has the following matrix

    \[M(\varphi)_{\mathcal{A}}^{\mathcal{A}}=\left[\begin{array}{ccccc}\cos \alpha&-\sin \alpha&0&\ldots &0\\\sin\alpha&\cos \alpha&0&\ldots&0\\0&0&1&\ldots&0\\&&\ldots&&\ldots&\\0&0&0&\ldots&1\end{array}\right]\]

where \mathcal{A} consists of an orthonormal basis of W after which we have an orthonormal basis of W^\bot. A rotation is an isometry.

It is easy to show that every isometry of a linear euclidean two-dimensional space is either a perpendicular reflection or a rotation.

Isometries of affine euclidean spaces

A mapping f\colon H\to M on affine euclidean spaces is called an isometry if one of the following equivalent conditions hold

  • f is an affine isomorphism and f'\colon T(H)\to T(M) is a isometry of linear spaces,
  • f is an affine mapping and f' is orthogonal in orthonormal bases of T(H) and T(M),

One can prove that f\colon H\to H is an isometry iff it preserves the distance between points , i.e. d(p,q)=d(f(p),f(q)) for any p,q\in H.

Mapping preserving measure of parallelepipeds

Notice that if R\subseteq H is a n-dimensional parallelepiped and f\colon H\to M is an affine isomorphism, then f[R] is also a n-dimensional parallelepiped.

Moreover, if R is a n-dimensional parallelepiped in a n-dimensional space, then \mu_n(f[R])=|\det f'|\cdot \mu_n(R). Indeed, if R=R(p_0;v_1,\ldots, v_n), then v_1,\ldots,v_n is a basis \mathcal{A}, and f'(v_1),\ldots, f'(v_n) is a basis \mathcal{A}'. Then for an orthonormal basis \mathcal{B}, we get

    \[\mu_n(f[R])=\sqrt{W(f'(v_1),\ldots, f'(v_n))}=|\det M(id)_{\mathcal{A}'}^{\mathcal{B}}|=|\det M(f)_{\mathcal{A}}^{\mathcal{B}}|=\]

    \[=|\det M(f)_{\mathcal{B}'}^{\mathcal{B}} M(id)_{\mathcal{A}}^{\mathcal{B}}|=|\det f'|\sqrt{W(v_1,\ldots, v_n)}=|\det f'|\cdot\mu_n(R).\]

Thus, an affine isomorphism f on n-dimensional affine euclidean spaces preserves measure of n-dimensional parallelepipeds if and only if \det f'=\pm 1.

Self-adjoint mappings

An endomorphism of a linear Euclidean space \varphi\colon V\to V is self-adjoint, if for any v,w\in V, \langle v,\varphi(w)\rangle = \langle \varphi(v),w\rangle.

Notice that then if \mathcal{A} is an orthonormal basis of V, the matrix M(\varphi)_{\mathcal{A}}^{\mathcal{A}} is symmetrical! Indeed, it is so because \langle v_i, \varphi(v_j)\rangle is the i-th coordinate of \varphi(v_j) with respect to \mathcal{A}. Reversly, if for an orthonormal basis \mathcal{A}, matrix M(\varphi)_{\mathcal{A}}^{\mathcal{A}} is symmetrical, then \varphi is self-adjoint.

It is easy to notice, that if v is an eigenvector and w\bot v, then \varphi(w)\bot v. Moreover, if v,w are eigenvectors for different eigenvalues, then v\bot w. Indeed if those values are a\neq b, then

    \[b\langle v,w\rangle =\langle v,bw\rangle=\langle av,w\rangle = a\langle v,w\rangle,\]

so \langle v,w\rangle= 0.

Also it can be proved every eigenvalue of a symmetrical matrix is real! Moreover, by an easy inductive argument using the fact of preserving perpendicularity to an eigenvector, we can prove that every symmetrical matrix (thus every self-adjoint mapping) has a basis consisting of eigenvectors. This basis is orthogonal, and can be made orthonormal. In other words, for every symmetrical matrix A there exists an orthogonal matrix C, such that C^TAC is diagonal!