What characterizes a normal operator on a finite-dimensional complex inner product space?

It satisfies $A^*A=AA^*$; equivalently, it has a unitary diagonalization.

What spectral description applies to a compact self-adjoint operator?

The space decomposes into eigenspaces, and the operator has a finite or countable spectral sum.

How can the spectrum differ from the eigenvalues?

A spectral value need not be an eigenvalue; multiplication by $t$ on $L^2([0,1])$ has spectrum $[0,1]$ but no nonzero eigenvectors.

What is a cyclic vector, and what does it imply for a bounded self-adjoint operator?

A vector whose iterated images under the operator span a dense subspace; its existence gives a scalar $L^2$ model with simple spectrum.

How is $f(A)$ represented in a spectral multiplication model?

If $A$ acts by multiplication by $\lambda$, then $f(A)$ acts by multiplication by $f(\lambda)$.

Linear algebraLesson 1 of 1 in the course Eigenvalues and eigenvectors

Spectral Theorem: From Hermitian Matrices to Operator Calculus

5 flashcards · 5 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

This lesson introduces the finite-dimensional spectral theorem for Hermitian operators. It explains why Hermitian operators have real eigenvalues and how their eigenvectors form an orthonormal basis.

Let $V$ be a finite-dimensional complex inner product space, and let $A:V\to V$ be linear. The operator is Hermitian (or self-adjoint in finite dimensions) when

All flashcards

What characterizes a normal operator on a finite-dimensional complex inner product space?: It satisfies $A^*A=AA^*$ ; equivalently, it has a unitary diagonalization.
What spectral description applies to a compact self-adjoint operator?: The space decomposes into eigenspaces, and the operator has a finite or countable spectral sum.
How can the spectrum differ from the eigenvalues?: A spectral value need not be an eigenvalue; multiplication by $t$ on $L^2([0,1])$ has spectrum $[0,1]$ but no nonzero eigenvectors.
What is a cyclic vector, and what does it imply for a bounded self-adjoint operator?: A vector whose iterated images under the operator span a dense subspace; its existence gives a scalar $L^2$ model with simple spectrum.
How is $f(A)$ represented in a spectral multiplication model?: If $A$ acts by multiplication by $\lambda$ , then $f(A)$ acts by multiplication by $f(\lambda)$ .

Quiz questions

Which statement characterizes a finite-dimensional normal operator on a complex inner product space?Answer: It can be diagonalized using an orthonormal basis of eigenvectors.
For a Hermitian matrix, how are its singular values related to its eigenvalues?Answer: Each singular value is the absolute value of an eigenvalue.
For $[Af](t)=tf(t)$ on $L^2([0,1])$ , which statement is correct?Answer: Its spectrum is $[0,1]$ , but it has no nonzero eigenvectors.
If $A$ acts by multiplication by $\lambda$ in a spectral representation, how does $f(A)$ act?Answer: It multiplies by $f(\lambda)$ .
For multiplication by $t$ on $L^2(\mathbb{R})$ , which condition describes the operator domain?Answer: The domain consists of functions $f$ for which $tf$ is square-integrable.

Based on the Wikipedia article Spectral theorem, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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