Question

1) Define an eigenvalue. 2) Show that if \(\mathrm{u}\) and \(\mathrm{v}\) are eigenvectors of a linear operator \(\mathrm{f}\) which belong to \(\lambda\) and if a is a real number, then (a) \(\mathrm{u}+\mathrm{v}\) and (b) au are also eigenvectors of \(\mathrm{f}\) which belong to \(\lambda\).

Short answer

An eigenvalue \(\lambda\) is a scalar value associated with a linear operator \(\mathrm{f}\) such that \(\mathrm{f}(\mathrm{v}) = \lambda \mathrm{v}\) for a nonzero vector \(\mathrm{v}\). If \(\mathrm{u}\) and \(\mathrm{v}\) are eigenvectors of \(\mathrm{f}\) belonging to \(\lambda\), and \(a\) is a real number, then we can show that (a) \(\mathrm{u}+\mathrm{v}\) and (b) \(a\mathrm{u}\) are also eigenvectors of \(\mathrm{f}\), belonging to the eigenvalue \(\lambda\). This is because \(\mathrm{f}(\mathrm{u}+\mathrm{v}) = \mathrm{f}(\mathrm{u})+\mathrm{f}(\mathrm{v}) = \lambda \mathrm{u} + \lambda \mathrm{v} = \lambda (\mathrm{u}+\mathrm{v})\) and \(\mathrm{f}(a\mathrm{u}) = a\mathrm{f}(\mathrm{u}) = a(\lambda \mathrm{u}) = \lambda (a\mathrm{u})\).

Step-by-step solution

Define an Eigenvalue

An eigenvalue, denoted by the symbol \(\lambda\), is a scalar value associated with a linear operator \(\mathrm{f}\) acting on a vector space such that for a nonzero vector \(\mathrm{v}\), the action of \(\mathrm{f}\) on \(\mathrm{v}\) only flattens or stretches the vector \(\mathrm{v}\), without changing its direction. Mathematically, this can be expressed as: \[\mathrm{f}(\mathrm{v}) = \lambda \mathrm{v}\] Here, \(\mathrm{v}\) is called an eigenvector of \(\mathrm{f}\) corresponding to the eigenvalue \(\lambda\).

Show that the sum of two eigenvectors is an eigenvector

We are given that \(\mathrm{u}\) and \(\mathrm{v}\) are eigenvectors of \(\mathrm{f}\), belonging to the eigenvalue \(\lambda\). This means: \[\mathrm{f}(\mathrm{u}) = \lambda \mathrm{u}\] \[\mathrm{f}(\mathrm{v}) = \lambda \mathrm{v}\] Now, let's consider the sum of these two eigenvectors, \(\mathrm{w} = \mathrm{u} + \mathrm{v}\). We will show that \(\mathrm{w}\) is also an eigenvector of \(\mathrm{f}\), belonging to the same eigenvalue \(\lambda\). Let's apply the linear operator \(\mathrm{f}\) to \(\mathrm{w}\): \[\mathrm{f}(\mathrm{w}) = \mathrm{f}(\mathrm{u} + \mathrm{v})\] Using the properties of linear operators, we can rewrite the expression as: \[\mathrm{f}(\mathrm{u} + \mathrm{v}) = \mathrm{f}(\mathrm{u}) + \mathrm{f}(\mathrm{v})\] From the given, we know that \(\mathrm{f}(\mathrm{u}) = \lambda \mathrm{u}\) and \(\mathrm{f}(\mathrm{v}) = \lambda \mathrm{v}\). Therefore, we can substitute these expressions to get: \[\lambda \mathrm{u} + \lambda \mathrm{v} = \lambda (\mathrm{u} + \mathrm{v})\] This shows that \(\mathrm{w} = \mathrm{u} + \mathrm{v}\) is also an eigenvector of \(\mathrm{f}\), belonging to the same eigenvalue \(\lambda\).

Show that the scalar multiple of an eigenvector is an eigenvector

We are given that \(\mathrm{u}\) is an eigenvector of \(\mathrm{f}\) belonging to the eigenvalue \(\lambda\), and \(a\) is a real number. Now, we want to show that \(a\mathrm{u}\) is also an eigenvector of \(\mathrm{f}\), belonging to the same eigenvalue \(\lambda\). Let's apply the linear operator \(\mathrm{f}\) to \(a\mathrm{u}\): \[\mathrm{f}(a\mathrm{u})\] Using the properties of linear operators, we can rewrite this expression as: \[a\mathrm{f}(\mathrm{u})\] From the given, we know that \(\mathrm{f}(\mathrm{u}) = \lambda \mathrm{u}\). Therefore, we can substitute this expression to get: \[a(\lambda \mathrm{u}) = \lambda (a\mathrm{u})\] This shows that \(a\mathrm{u}\) is also an eigenvector of \(\mathrm{f}\), belonging to the same eigenvalue \(\lambda\).

Key concepts

Linear Operator

A linear operator is a mathematical entity that operates on vectors in a vector space. It is denoted by \( \mathrm{f} \) and adheres to two principal properties: additivity and homogeneity.
These properties enable the linear operator to distribute over addition and scalar multiplication, ensuring consistency and predictability in operations.

• **Additivity**: For any two vectors \( \mathrm{u} \) and \( \mathrm{v} \), a linear operator satisfies \( \mathrm{f}(\mathrm{u} + \mathrm{v}) = \mathrm{f}(\mathrm{u}) + \mathrm{f}(\mathrm{v}) \).
• **Homogeneity**: For any vector \( \mathrm{v} \) and scalar \( a \), it holds that \( \mathrm{f}(a \cdot \mathrm{v}) = a \cdot \mathrm{f}(\mathrm{v}) \).

These properties make linear operators crucial in simplifying complex vector operations. Linear operators are widely used, not only in mathematics, but also in engineering and physics to model real-world phenomena that maintain linear relationships.

Eigenvector

An eigenvector is an essential concept in linear algebra. It is a special type of vector, denoted typically as \( \mathrm{v} \), which remains aligned in its direction even after a linear transformation is applied to it. This transformation is expressed through a linear operator \( \mathrm{f} \).
Eigenvectors are non-zero vectors that satisfy a seminal relationship:

• **Eigen Equation**: \( \mathrm{f}(\mathrm{v}) = \lambda \mathrm{v} \), where \( \lambda \) is the eigenvalue.

To understand this better, imagine pointing a laser beam in space where it travels along a specific line. Even if you apply different lenses that alter its intensity (think of it as scaling by \( \lambda \)), the beam direction remains unchanged.
Eigenvectors are particularly meaningful in systems where directionality is important, such as in computer graphics, quantum mechanics, and data analysis in statistics.

Vector Space

Vector spaces lay the foundational framework for exploring vectors and their interactions. A vector space consists of a collection of objects called vectors, along with operations like addition and scalar multiplication that satisfy specific rules.
These spaces provide an environment where mathematical concepts like eigenvalues and eigenvectors can be clearly defined and analyzed. The basic properties that define vector spaces are:

• **Closure**: For any two vectors in the space, their sum must also be in the same space.
• **Associativity**: Vector addition must obey the associative rule: \( (\mathrm{u} + \mathrm{v}) + \mathrm{w} = \mathrm{u} + (\mathrm{v} + \mathrm{w}) \).
• **Existence of a zero vector**: There exists a zero vector \( \mathbf{0} \) such that any vector plus this zero vector remains unchanged: \( \mathrm{v} + \mathbf{0} = \mathrm{v} \).
• **Distributive property**: For any scalar \( a \) and vectors \( \mathrm{u} \) and \( \mathrm{v} \), the properties \( a(\mathrm{u} + \mathrm{v}) = a\mathrm{u} + a\mathrm{v} \) and \( (a + b)\mathrm{v} = a\mathrm{v} + b\mathrm{v} \) hold.

Vector spaces are ubiquitous in various domains of mathematics and science, such as function spaces in calculus and state spaces in quantum mechanics.

Scalar Multiplication

Scalar multiplication is a straightforward yet influential operation in vector spaces. It involves multiplying a vector by a scalar value, effectively scaling the vector. Here, \( a \) represents a scalar, and \( \mathrm{v} \) is a vector. The operation \( a \mathrm{v} \) results in a new vector that is collinear with \( \mathrm{v} \), but scaled by \( a \).
The main properties of scalar multiplication are:

• **Associativity of scalar and vector**: \((ab) \mathrm{v} = a(b \mathrm{v}) \) for scalars \( a \) and \( b \).
• **Identity element of scalar 1**: \( 1\mathrm{v} = \mathrm{v} \).
• **Distributed over scalar addition**: \( (a + b)\mathrm{v} = a\mathrm{v} + b\mathrm{v} \).

Scalar multiplication is vital in scaling eigenvectors and demonstrating properties like the span and directionality changes in transformations. Its simplicity belies its utility and significance in linear transformations and vector space computations. Understanding how vectors change under scalar multiplication provides insight into the behavior of solutions in linear differential equations and simulations in physics and engineering.