Raising the bar Orthogonal matrices: definition, properties, relationship to transition matrix

Aug 18

Orthogonal matrices

Orthogonal matrices

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

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

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

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

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

Since the origin is unchanged under any linear mapping, $A0=0,$ 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 $\det A=1$ we have rotation and in case $\det A=-1$ - rotation combined with reflection.

Another interpretation is suggested by the next exercise.

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

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

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

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

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