Orthogonal Matrices

Date: 01/07/98 at 23:54:34
From: Thomas Dehler
Subject: orthogonal matrices

If A is an orthogonal matrix, then the determinant of A is either 1 
or -1.

How do I prove this?

Date: 01/09/98 at 10:03:03
From: Doctor Joe
Subject: Re: orthogonal matrices

Hi Thomas, 

The problem you have asked belongs to a set of just about the nicest 
set of problems in linear algebra. I hope you already know the 
following theorems and I can use them without proofs:

Theorem A

Let A and B be any two square matrices with the same dimension.
Denoting the determinant of a matrix A by det A, (likewise for B),
we have:
             det AB  =  det A  *  det B.

In other words, the determinant preserves the product of matrices.

(Remark: This gives an insight to why the determinant gives rise to 
many homomorphisms between certain subgroups of square matrices and 
the non-zero real number line.)

Theorem B

Denote the transpose of A as A^T for a given square matrix A. The 
determinant of A^T is the same as the determinant of A itself; i.e. 
det A^T = det A.

Now, let's prove your statement:

"If A is orthogonal, then det A = either 1 or -1."

Proof:

By definition, A is orthogonal means that A^T = A^(-1)  (read as the 
transpose of A is equal to its inverse).

Of course, from the definition of inverse of a matrix,

              A * A^(-1) = I,

where I denotes the usual identity matrix.

It follows that for this orthogonal matrix A,

              A * A^T    = I.

Now taking determinant on both sides,

           det (A * A^T) = det I

Then, invoking Thm A, det A * det A^T = 1 (since det I = 1. Recall 
that the determinant of a diagonal matrix is the product of its 
diagonal terms.)

Next, invoking Thm B, det A * det A = 1 (since det A^T = det A).

It follows that (det A)^2 = 1.  Hence, det A = 1 or -1.  (proven)

Remarks.

When restricted to the case of A being dimension 2, an orthogonal 
matrix is some rotation or reflection. (As an exercise, try to find 
out exactly which rotation and which reflection.)

Thus, det of such matrices are one (they are area-preserving).

-Doctor Joe,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/