Orthogonal and Unitary Matrices

Getting to Know Orthogonal and Unitary Matrices

Orthogonal and unitary matrices are very special types of matrices. Imagine them as "clean" transformations that don't change distances and angles in space, only rotating or reflecting objects.

The difference is simple. Orthogonal matrices work with real numbers, while unitary matrices work with complex numbers. Both have the same properties, just different versions.

Mathematical Definitions

Orthogonal Matrices

A square real matrix $A \in \mathbb{R}^{n \times n}$ is called orthogonal if:

A^{-1} = A^T

This means, to get the inverse of this matrix, we just transpose it. Very practical, right?

This is equivalent to:

A^T A = A A^T = I

Unitary Matrices

A square complex matrix $A \in \mathbb{C}^{n \times n}$ is called unitary if:

A^{-1} = A^H

Here $A^H$ is the conjugate transpose of $A$ . The concept is similar, just for complex numbers.

This is also equivalent to:

A^H A = A A^H = I

Real orthogonal matrices are actually a special case of unitary matrices, since $\mathbb{R}^{n \times n} \subset \mathbb{C}^{n \times n}$ .

Interesting Determinant Properties

What's interesting about orthogonal and unitary matrices is that their determinants always have absolute value 1. Why is this?

For a unitary matrix $A$ , we have $A^H A = I$ . If we calculate its determinant:

1 = \det I = \det(A^H A) = \det A^H \cdot \det A = \overline{\det A} \cdot \det A = |\det A|^2

So $|\det A| = 1$ . For orthogonal matrices, the proof is the same, just using $A^T A = I$ .

Special Eigenvalues

Eigenvalues of orthogonal and unitary matrices also have special properties. Every eigenvalue $\lambda$ always satisfies:

|\lambda| = 1

Why is this? Suppose $A \cdot v = \lambda \cdot v$ for an eigenvector $v \neq 0$ . For the complex case, we can calculate:

v^H v = v^H A^H A v = (A \cdot v)^H (A \cdot v) = (\lambda \cdot v)^H (\lambda \cdot v)

= \overline{\lambda} \cdot \lambda \cdot v^H v = |\lambda|^2 \cdot v^H v

Since $v^H v \neq 0$ , then $|\lambda|^2 = 1$ , so $|\lambda| = 1$ .

Forms of Eigenvalues

For real orthogonal matrices, the eigenvalues can be 1 or -1 if real. But if complex, they can be written as:

\lambda = \exp(i\varphi) = \cos \varphi + i \sin \varphi

This means complex eigenvalues lie on the unit circle in the complex plane.

Concrete Example of Rotation Matrix

Let's look at a familiar example, the rotation matrix:

A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}

We can check that this is an orthogonal matrix:

A^T A = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}

= \begin{pmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha(-\sin \alpha) + \sin \alpha \cos \alpha \\ (-\sin \alpha) \cos \alpha + \cos \alpha \sin \alpha & (-\sin \alpha)(-\sin \alpha) + \cos^2 \alpha \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Finding Eigenvalues

The characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} \cos \alpha - t & -\sin \alpha \\ \sin \alpha & \cos \alpha - t \end{pmatrix}

= (\cos \alpha - t)^2 + \sin^2 \alpha = \cos^2 \alpha + \sin^2 \alpha - 2t \cos \alpha + t^2

= 1 - 2t \cos \alpha + t^2

The eigenvalues are:

\lambda_{1,2} = \cos \alpha \pm \sqrt{\cos^2 \alpha - 1} = \cos \alpha \pm \sqrt{-\sin^2 \alpha} = \cos \alpha \pm i \sin \alpha

The result is $\lambda_{1,2} = e^{\pm i\alpha}$ .

The transformation $\mathbb{R}^2 \to \mathbb{R}^2 : x \mapsto A \cdot x$ represents a rotation by angle $\alpha$ . For $\alpha \neq 0$ and $\alpha \neq \pi$ , this matrix has no real eigenvalues, but has two complex eigenvalues.

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