Eigenvalue

Definition: Eigenvalue

Let $A\in F^{n\times n}$ be a square matrix.

We say that $\lambda \in F$ is an eigenvalue of $A$ if there is a non-zero vector $\vec{v}\in F^n$ such that

$$A\vec{v}=\lambda \vec{v}$$

In this case, we also say that $\lambda$ has the eigenvector $\vec{v}$.

NOTE

An eigenvalue can have multiple eigenvectors.

Theorem: Count of Eigenvalues

A square matrix $A\in F^{n\times n}$ has at most $n$ different eigenvalues.

PROOF

TODO

Theorem: Algebraic and Geometric Multiplicity

The geometric multiplicity and the algebraic multiplicity of every eigenvalue $\lambda$ of a square matrix $A\in F^{n\times n}$ obey the following inequality:

$$1\le \mathrm{geo}(\lambda )\le \mathrm{alg}(\lambda )$$

PROOF

TODO

Theorem: Sum of the Eigenvalues

The distinct eigenvalues ${\lambda }_1,\cdots ,{\lambda }_r$ of a square matrix $A\in F^{n\times n}$ and their algebraic multiplicities can be used to calculate the trace of $A$:

$$\sum _{k=1}^r{\lambda }_k\cdot \mathrm{alg}({\lambda }_k)=\mathrm{Tr}(A)$$

PROOF

TODO

Theorem: Product of the Eigenvalues

The distinct eigenvalues ${\lambda }_1,\cdots ,{\lambda }_r$ of a square matrix $A\in F^{n\times n}$ and their algebraic multiplicities can be used to calculate the determinant of $A$:

$$\prod _{k=1}^r{\lambda }_k^{\mathrm{alg}({\lambda }_k)}=\det (A)$$

PROOF

TODO