Definition of Orthogonal Matrix: Case 1 or 2?

In summary, the definition of an orthogonal matrix is a matrix where all rows are orthonormal and all columns are orthonormal. If a matrix is diagonalizable, then its eigenvectors are all distinct and orthogonal, and by only normalizing its columns, we guarantee that its rows are also orthonormal.
  • #1
sjeddie
18
0
Is the definition of an orthogonal matrix:

1. a matrix where all rows are orthonormal AND all columns are orthonormal

OR

2. a matrix where all rows are orthonormal OR all columns are orthonormal?

On my textbook it said it is AND (case 1), but if that is true, there's a problem:
Say we have a square matrix A, and we find its eigenvectors, they are all distinct so A is diagonalizable. We put the normalized eigenvectors of A as the columns of a matrix P, and (our prof told us) P becomes orthogonal and P^-1 = P^T. My question is how did P become orthogonal straight away? By only normalizing its columns how did we guarantee that its rows are also orthonormal?
 
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  • #2
It turns out that the rows of a square matrix are orthonormal if and only if the columns are orthonormal. Another way to express that the condition that all columns are orthonormal is that [tex]A^T A = I[/tex] (think about why this is). Then we see that if [tex]v \in \mathbb{R}^n[/tex], [tex]\parallel x \parallel^2 = x^T x = x^T ( A^T A ) x = ( A x )^T ( A x ) = \parallel A x \parallel^2[/tex], and therefore A is injective. Since we are working with finite-dimensional spaces, A must also be surjective, so for [tex]v \in \mathbb{R}^n[/tex], there exists [tex]w \in \mathbb{R}^n[/tex] with v = Aw, and therefore [tex]A A^T v = A A^T A w = A w = v[/tex], so [tex]A A^T = I[/tex] as well. You can check this this implies that the rows of A are orthonormal. The proof of the converse is similar.

Note that this argument relies on the finite-dimensionality of our vector space. If you move up to infinite dimensional spaces, there may be transforms T with [tex]T^*T = I[/tex] but [tex]T T^* \neq I[/tex]. This type of behavior is what makes functional analysis and operator algebras fun! :smile:
 
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  • #3
Theres actually an easier way to see that [tex]A^T A = I[/tex] implies A is injective, I just tend to think in terms of isometries like I wrote. If v is such that Av = 0, then [tex]0 = A^T 0 = A^T A v = v[/tex], so A is injective. Some may prefer this purely algebraic argument.
 
  • #4
Ah I see, thank you rochfor1, the (A^T)(A) = I thing makes a lot of sense :)
 

Related to Definition of Orthogonal Matrix: Case 1 or 2?

What is an orthogonal matrix?

An orthogonal matrix is a square matrix where the rows and columns are orthogonal unit vectors, meaning they are mutually perpendicular and have a length of 1. In other words, the dot product of any two rows or columns is 0.

What is the difference between Case 1 and Case 2 in the definition of an orthogonal matrix?

In Case 1, the orthogonal matrix has real numbers as its entries. In Case 2, the orthogonal matrix has complex numbers as its entries. Both cases have the same properties of orthogonality and unit length, but they differ in the type of numbers used.

How can an orthogonal matrix be used in linear transformations?

An orthogonal matrix can be used to perform rotations, reflections, and other transformations of a vector or set of vectors without changing their length.

What is the inverse of an orthogonal matrix?

The inverse of an orthogonal matrix is its transpose, meaning that the rows become columns and the columns become rows. This is because the dot product of two orthogonal vectors is 0, so the transpose of an orthogonal matrix will also have this property.

What are some applications of orthogonal matrices?

Orthogonal matrices are used in many areas of science and engineering, such as computer graphics, signal processing, and quantum mechanics. They have applications in data compression, image processing, and error correction, among others.

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