28
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. 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 - 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.

Proof. Orthonormality means that u_i^Tu_j=1 if i=j and u_i^Tu_j=0, if i\neq j. This implies 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.

Remark. Equation (1) shows that columns of an orthogonal matrix constitute an orthonormal basis. The equation UU^T=I shows that the same is true for rows.

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