Eigenvalues

Definition

To be done. eigenvectors. Matrices and operators

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Determination by the trace

Importantly, the knowledge of the trace of all the powers of a matrix let us obtain the eigenvalues according to this. It is due to the Newton-Girard identities.

Invariance by similarity transformation

Let $A$ and $B$ be two operators on a Hilbert space $H$ such that $B = P^{-1}AP$ for some invertible operator $P$ on $H$ (i.e., $A$ and $B$ are related by a similarity transformation). We wish to show that $A$ and $B$ have the same eigenvalues.

First, let's take $\lambda$ as an eigenvalue of $A$ with an associated eigenvector $|v\rangle$, i.e.

$$ A|v\rangle = \lambda|v\rangle $$

$$ P^{-1}A|v\rangle = \lambda P^{-1}|v\rangle $$

$$ BP^{-1}|v\rangle = \lambda P^{-1}|v\rangle $$

This equation shows that $P^{-1}|v\rangle$ is an eigenvector of $B$ with eigenvalue $\lambda$. Thus, any eigenvalue of $A$ is also an eigenvalue of $B$. Now we can apply a symmetric argument.

Complex case

See complex eigenvalues

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es