Problem 11 - Differential Equations and Linear Algebra

Question

Show that the given set of vectors is an orthogonal set in \(\mathbb{C}^{n}\), and hence obtain an orthonormal set of vectors in \(\mathbb{C}^{n}\) in each case. $$\\{(1-i, 3+2 i),(2+3 i, 1-i)\\}$$

Step-by-Step Solution

Answer

The given set of vectors is orthogonal, as their dot product is equal to 0. The orthonormal set of vectors in \(\mathbb{C}^{n}\) is: $$\\{ (\frac{1-i}{\sqrt{15}}, \frac{3+2i}{\sqrt{15}}), (\frac{2+3i}{\sqrt{15}}, \frac{1-i}{\sqrt{15}}) \\}$$

Step 1: Compute the dot product of the given vectors

Let the given vectors be: $$A = (1-i, 3+2i)$$ $$B = (2+3i, 1-i)$$ Compute the dot product, also known as the inner product, of vectors A and B in complex space, which can be computed as: $$\langle A, B \rangle = \sum_{k=1}^{n} A_k \cdot \overline{B_k}$$ In this case: $$\langle A, B \rangle = (1-i)(2-3i) + (3+2i)(1+i)$$

Step 2: Simplify the dot product expression

Simplify the terms in the expression: $$\langle A, B \rangle = (1-i)(2-3i) + (3+2i)(1+i) = (2 - 3i - 2i + 3) + (3 + 2i + 3i - 2)$$

Step 3: Combine like terms

Combine the terms with real parts and the terms with imaginary parts: $$\langle A, B \rangle = (2+3) + (-3i -2i) + (3-2) + (2i+3i) = 5 + 0i = 0$$ Since the dot product of the two vectors is equal to 0, we can conclude that the given set of vectors is orthogonal.

Step 4: Obtain the orthonormal set of vectors

Now, we need to normalize the orthogonal vectors to obtain an orthonormal set. To normalize a vector, divide each component of the vector by its magnitude. Compute the magnitude of the two vectors: $$||A|| = \sqrt{(1-i)(1+i) + (3+2i)(3-2i)} = \sqrt{1+1^2 + 3^2+2^2} = \sqrt{15}$$ $$||B|| = \sqrt{(2+3i)(2-3i) + (1-i)(1+i)} = \sqrt{2^2+3^2 + 1+1^2} = \sqrt{15}$$ Normalize the vectors: $$A' = \frac{1}{\sqrt{15}}(1-i, 3+2i) = \frac{1}{\sqrt{15}}(1-i) + \frac{1}{\sqrt{15}}(3+2i)$$ $$B' = \frac{1}{\sqrt{15}}(2+3i, 1-i) = \frac{1}{\sqrt{15}}(2+3i) + \frac{1}{\sqrt{15}}(1-i)$$ The orthonormal set of vectors in \(\mathbb{C}^{n}\) is: $$\\{ (\frac{1-i}{\sqrt{15}}, \frac{3+2i}{\sqrt{15}}), (\frac{2+3i}{\sqrt{15}}, \frac{1-i}{\sqrt{15}}) \\}$$

Key Concepts

Complex Vector Space

A complex vector space is a fundamental concept in linear algebra, where vectors are allowed to have complex numbers as their components. This differs from real vector spaces, where vector components are strictly real numbers. In a complex vector space, operations such as vector addition and scalar multiplication adhere to the rules of complex arithmetic.

When we talk about a complex vector space like \( \mathbb{C}^n \), we refer to an \( n \)-dimensional space where each vector component can be a complex number.

• A complex number is expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with the property \( i^2 = -1 \).
• Complex arithmetic involves operations such as addition, subtraction, multiplication, and division, which extend naturally from operations with real numbers.

The concept of complex vector spaces allows us to generalize many ideas from standard calculus and linear algebra to more complex domains, making them especially useful in fields like quantum mechanics, signal processing, and more.

Inner Product

The inner product is a generalization of the dot product to complex vector spaces and plays a crucial role in determining the angles and lengths of vectors within those spaces. It is especially important when dealing with orthogonal or orthonormal sets of vectors.

For complex vectors, the inner product is given by:\[ \langle A, B \rangle = \sum_{k=1}^{n} A_k \cdot \overline{B_k} \]This formula is similar to the dot product in real spaces, but with an essential twist: the complex conjugate of each component of the second vector is used. The complex conjugate, denoted as \( \overline{B_k} \), is the number \( \overline{a + bi} = a - bi \).

• The inner product of a vector with itself helps define a vector's magnitude (or norm) in complex spaces.
• Vectors are orthogonal if their inner product equals zero, meaning they meet at a right angle, essential for concepts like orthogonality.

Understanding the inner product in complex spaces is fundamental to grasping the nature of vector relationships and dimensions in areas like quantum computing.

Orthonormal Set

An orthonormal set is a collection of vectors in a vector space that are both orthogonal and normalized. "Orthogonal" means that each pair of different vectors has an inner product of zero, signifying they are at right angles to each other. "Normalized" means each vector has a magnitude of 1.

In practice, to convert an orthogonal set into an orthonormal set, each vector is divided by its magnitude:- **Magnitude Calculation:** The magnitude is found using the inner product: \( ||A|| = \sqrt{\langle A, A \rangle} \).- **Normalization:** Once magnitudes are determined, normalize by dividing each vector by its magnitude to ensure all vectors have a unit length.

• An orthonormal set in a complex vector space like \( \mathbb{C}^n \) is particularly useful because it simplifies many mathematical operations.
• They maintain nice properties under transformations and are pivotal in methods like the Gram-Schmidt process in linear algebra.

The importance of orthonormal sets stems not only from their mathematical elegance but their practical applications in signal processing and the representation of complex data structures.