Raising the bar

Orthogonal matrices

Definition 1. A square matrix $latex A$ is called orthogonal if $latex A^TA=I.$

Exercise 1. Let $latex A$ be orthogonal. Then a) $latex A^{-1}=A^T,$ b) the transpose $latex A^T$ is orthogonal, c) the inverse $latex A^{-1}$ is orthogonal, d) $latex |\det A|=1.$

Proof. a) $latex A^T$ is the left inverse of $latex A.$ Hence, $latex A$ is invertible and its inverse is $latex A^T.$ b) $latex AA^T=I$ from the inverse definition. Part c) follows from parts a) and b). d) Just apply $latex \det$ to the definition to get $latex (\det A)^{2}=1.$

Exercise 2. An orthogonal matrix preserves scalar products, norms and angles.

Proof. For any vectors $latex x,y$ scalar products are preserved: $latex (Ax)\cdot(Ay)=(A^TAx)\cdot(y)=x\cdot y.$ Therefore vector lengths are preserved: $latex \|Ax\| =\|x\|.$ Cosines of angles are preserved too, because $latex \frac{(Ax)\cdot(Ay)}{\|Ax\|\|Ay\|}=\frac{x\cdot y}{\|x\|\| y\|}.$ Thus angles are preserved.

Since the origin is unchanged under any linear mapping, $latex A0=0,$ Exercise 2 gives the following geometric interpretation of an orthogonal matrix: it is rotation around the origin (angles and vector lengths are preserved, while the origin stays in place). Strictly speaking, in case $latex \det A=1$ we have rotation and in case $latex \det A=-1$ - rotation combined with reflection.

Another interpretation is suggested by the next exercise.

Exercise 3. If $latex u_1,...,u_n$ is an orthonormal basis, then the matrix $latex U=(u_1,...,u_n)$ is orthogonal. Conversely, rows or columns of an orthogonal matrix form an orthonormal basis.

Proof. Orthonormality means that $latex u_i^Tu_j=1$ if $latex i=j$ and $latex u_i^Tu_j=0,$ if $latex i\neq j.$ These equations are equivalent to orthogonality of $latex U:$

(1) $latex U^TU=\left(\begin{array}{c}u_1^T \\... \\u_n^T\end{array}\right)\left(u_1,...,u_n\right)=I.$

Exercise 4. Let $latex u_1,...,u_n$ and $latex v_1,...,v_n$ be two orthonormal bases. Let $latex A=V^{-1}U$ be the transition matrix from coordinates $latex \xi$ in the basis $latex u_1,...,u_n$ to coordinates $latex \xi^\prime$ in the basis $latex v_1,...,v_n$. Then $latex A$ is orthogonal.

Proof. By Exercise 3, both $latex U$ and $latex V$ are orthogonal. Hence, by Exercise 1 $latex V^{-1}$ is orthogonal. It suffices to show that a product of two orthogonal matrices $latex M,N$ is orthogonal: $latex (MN)^TMN=N^TM^TMN=N^TN=I.$

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