Chapter 13: Problem 24

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

Short Answer

Expert verified

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 by step solution

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 \(\| c v \| = |c| \cdot \| v \|\).

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 \(A\).

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.