Problem 24 - A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry

Question

Show that a non-zero vector \(u\) is an cigenvector of an operator \(A\) if and only if \(|\langle u \mid A u\rangle|=\| A u|| u \mid .\)

Step-by-Step Solution

Answer

The vector \(u\) is an eigenvector of the operator \(A\) if and only if \(|\langle u \mid A u\rangle|=\| A u|| u \|\). It was proven by assuming \(u\) is an eigenvector of \(A\), and then also showing the reverse - that this condition implies that \(u\) must be an eigenvector, demonstrating the 'if and only if' relationship.

Step 1: Understand the Meaning of Each Term

Firstly, it's necessary to keep in mind what each of the terms in the equation mean. An eigenvector is a non-zero vector that only changes by a scalar factor when a linear transformation is applied to it, expressed as \(Au=\lambda u\). The inner product \(\langle u \mid A u\rangle\) is the projection of \(A u\) onto \(u\). The norm of a vector \(v\), \(\| v \|\), is its length, and for a positive scalar \(c\), we have that \(\| c v \| = |c| \cdot \| v \|\).

Step 2: Show The 'If' Direction

We know that if \(u\) is an eigenvector of \(A\), there exists some scalar \(\lambda\) for which \(A u=\lambda u\). Hence, \(\langle u \mid A u\rangle = \langle u \mid \lambda u\rangle = \lambda \langle u \mid u \rangle\). Given that norm squared equals the inner product of a vector with itself \(\| u \|^2 = \langle u \mid u \rangle\), we can rewrite the previous expression as \(\langle u \mid A u\rangle = |\lambda| \| u \|^2\). Similarly, the norm of \(A u\) equals the absolute value of \(\lambda\) times the norm of \(u\), thus \(\| A u \|=\| \lambda u \|=|\lambda| \| u \|\). So, indeed, \(|\langle u \mid A u\rangle|=\| A u|| u \|\) if \(u\) is an eigenvector of \(A\).

Step 3: Show The 'Only If' Direction

Assume that \(|\langle u \mid A u\rangle|=\| A u|| u \|\), without knowing that \(u\) is an eigenvector of \(A\). This condition will hold only if \(A u = \lambda u\) for some scalar \(\lambda\). This means that \(u\) is an eigenvector of \(A\) and it proves the 'only if' direction.

Key Concepts

Linear Transformation

A linear transformation is an essential concept in linear algebra. It refers to a function between two vector spaces that preserves vector addition and scalar multiplication. When a vector is transformed by a matrix, it may undergo changes in direction and magnitude, except when it becomes an eigenvector. Here's a simple way to understand it:

• A linear transformation can be thought of as a rule applied to vectors that doesn't 'bend' or 'stretch' them in unpredictable ways.
• Mathematically, if you have a vector \(v\) and apply a linear transformation \(T\), the result is \(T(v)\), which is another vector.
• In matrix form: If \(A\) is a matrix representing the transformation, and \(x\) is a vector, then the transformation \(T(x)\) is represented as \(Ax\).

Understanding these transformations is key to grasping more complex concepts like eigenvectors, which remain unchanged in direction under these transformations.

Inner Product

The inner product is a fundamental operation that measures the "closeness" of two vectors. It is often referred to as the dot product when dealing with real vectors. In more abstract settings, it's a crucial concept because it helps define angles and lengths.

• The inner product of two vectors \(u\) and \(v\) is written as \(\langle u, v \rangle\) and computes as \(u_1v_1 + u_2v_2 + \dots + u_nv_n\).
• This operation results in a scalar and is a generalized way to think about projection.
• In context, when we consider \(\langle u \mid A u \rangle \), we're looking at how much of \(A u\) aligns with \(u\).

An understanding of the inner product is crucial in recognizing the relationship between vectors in transformations and checking if a vector is an eigenvector.

Vector Norm

The norm of a vector, often interpreted as the "length" or "magnitude," extends the notion of distance in vector spaces. Knowing how to compute and use it gives insight into vector size and behavior.

• The norm of a vector \(v\), denoted as \(\|v\|\), is calculated as \(\sqrt{v_1^2 + v_2^2 + \dots + v_n^2}\) for a vector \(v = (v_1, v_2, \dots, v_n)\).
• In the context of this problem, it's important to see how the norm relates to the inner product: \(\|v\|^2 = \langle v \mid v \rangle\).
• Understanding vector norms helps you gauge the scale of transformations, especially when a scalar multiplies a vector.

This concept is instrumental in understanding the conditions under which a vector is an eigenvector.

Scalar Factor

A scalar factor is a simple yet powerful idea. When you multiply a vector by a scalar, you scale its magnitude without altering its direction.

• A vector \(v\) multiplied by a scalar \(\lambda\) becomes \(\lambda v\). This operation scales the vector by \(\lambda\).
• If \(\lambda\) is positive, the vector maintains its direction; if \(\lambda\) is negative, the direction is reversed.
• Scalar factors are fundamental in eigenvalue problems, where you see expressions like \(Au = \lambda u\), conveying that \(u\) is scaled by \(\lambda\) under the transformation \(A\).

Understanding scalar multiplication helps you grasp how eigenvectors scale under linear transformations and why they are central to these algebraic investigations.