Equivalence of isometries

Orthogonal and symmetric maps: Isometries

Equivalence of isometries

The question when two linear isometries are equal up to a choice of basis, is answered by the following theorem, which is comparable to theorem Matrix equivalence for general linear maps. In this case, each pair of linear isometries between two inner product spaces of finite dimension is matrix equivalent, while the corresponding invertible matrices can be chosen orthogonal.

Matrix equivalence for linear isometries

Let #V# and #W# be finite-dimensional inner product spaces of dimension #m# and #n#, respectively. Suppose that #L# is a linear isometry #V\to W#.

We have #n\ge m#.
If #\alpha# is an orthonormal basis of #V#, then there is a basis #\beta# of #W# such that the matrix #{}_\beta L_\alpha# is equal to the #(n\times m)#-matrix \(\matrix{I_m\\ 0}\).
For each linear isometry #M:V\to W# there is an orthogonal map #Y:W\to W# such that #L=Y\,M#.

1. The proof of the first statement follows from the fact that #L# is injective. In particular, the kernel of #L# consists only of the zero vector, so the Rank-nullity theorem for linear maps implies that the image has dimension #m#; thus, #W# has dimension at least #m#, that is, #n\ge m#. This can also be seen by observing that the image of the orthonormal basis #\alpha=\basis{\vec{a}_1,\ldots,\vec{a}_m}# is an orthonormal, and therefore independent, system of vectors in #W#.

2. To prove the second statement we expand the images under #L# of the system #\alpha=\basis{\vec{a}_1,\ldots, \vec{a}_m}# to an orthonormal basis #\beta# of #W#. Thus, the basis #\beta# consists of the basis #\basis{L(\vec{a}_1),\ldots, L(\vec{a}_m)}# of #\im{L}# and an orthonormal basis of the orthogonal complement #\im{L}^\perp#. The matrix #{}_\beta L_\alpha# of #L# with respect to the bases #\alpha# and #\beta# is equal to the #(n\times m)#-matrix \(\matrix{I_m\\ 0}\).

3. We will now prove the third statement. The second statement implies that there is an orthonormal basis #\gamma# for #W# such that #{}_\gamma M_\alpha = \matrix{I_m\\ 0}#. In particular, we have \[{}_\beta L_\alpha= \matrix{I_m\\ 0} ={}_\gamma M_\alpha={}_\gamma I_\beta \,{}_\beta M_\alpha \] Let #Y:W\to W# be the linear map which is determined with respect to the basis #\beta# of #W# by the matrix #{}_\gamma I_\beta#. Then, it follows from the above equalities that #L = Y\, M#.

We still need to prove that #Y# is orthogonal. Because #\beta# and #\gamma# are orthonormal bases, the transition matrix is #{}_\gamma I_\beta# is orthogonal. Therefore, according to theorem Orthogonal maps on finite-dimensional spaces and orthonormal systems, also the linear map #Y:W\to W# determined by it is orthogonal. This proves the theorem.

Matrix equivalence In particular, the matrix of each isometry #V\to W# with respect to an orthonormal basis of #V# and an orthonormal basis of #W#, is matrix equivalent to the diagonal #(n\times m)#-matrix #\matrix{I_m\\ 0}#, where #0# represents the zero matrix of size #(n-m)\times m#.

Existence The converse of the first statement is also true: if #n\ge m#, there is a linear isometry #V\to W#. In order to see this, choose an orthonormal basis #\basis{\vec{a}_1,\ldots,\vec{a}_m}# of #V# and an orthonormal basis #\basis{\vec{b}_1,\ldots,\vec{b}_n}# of #W#. Then there is a unique linear map #L:V\to W# determined by #L(\vec{a}_i) = \vec{b}_i# for #i=1,\ldots,m#. This map is an isometry in view of the theorem Orthogonal maps in finite-dimensional spaces and orthonormal systems.

According to Matrix equivalence for linear isometries, the linear isometry #L:\mathbb{R}^2\to\mathbb{R}^3# given by the matrix \[A = 1\, \matrix{0 & 0 \\ 1 & 0 \\ 0 & 1 \\ }\] is matrix equivalent to the matrix \(J =\matrix{1 & 0 \\ 0 & 1 \\ 0 & 0 \\ }\).

There is even an orthogonal matrix #Y# such that #A = Y\, J#. Determine such a matrix.

#Y = # #\cdot \matrix{0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ }#

If we begin with the standard basis for #\mathbb{R}^2#, then the first two columns of #A# form an orthonormal system. We extend this pair with a vector to an orthonormal basis of #\mathbb{R}^3# and add this vector as a column to the matrix #A# in order to get the matrix #Y#:
\[ Y =\cdot \matrix{0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ }\] The matrix #A# consists of the first two columns of #Y# and thus can be written as \[A =1\, \matrix{0 & 0 \\ 1 & 0 \\ 0 & 1 \\ } =\cdot \matrix{0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ }\, \matrix{1 & 0 \\ 0 & 1 \\ 0 & 0 \\ }= Y\,J \]

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