The spectrum of $A$, $\Sigma (A)$, is the set of all eigenvalues matrix $A$ have. Some matrices do not have (real) eigenvalue, such as rotation matrix.
Note: Eigenvalues <-> Invertibility; Eigenvectors <-> Diagonalizability.
Fact: $f(\lambda)$ is an eigenvalue of $f(A)$ with eigenvector $\mathbf{x}$.
Lemma: If $S$ diagonalizes $A$, then it also diagonalizes $f(A)$.
Fact: Suppose $A$ is invertible, then $g \in \mathbb{C}[Z,Z^{-1}]$ (generalized series of A) works too. (i.e. $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$ .)
A family $\mathcal{F}$ of matrices is commutable, if $AB=BA, \forall A, B \in \mathcal{F}$.
A family $\mathcal{F}$ of matrices is simultaneously diagonalizable, if exist $S$, s.t. $S^{-1} A S$ is diagonal, $\forall A \in \mathcal{F}$.
A subspace $\mathcal{W}$ is $A$-invariant, if $Aw \in \mathcal{W}, \forall w \in \mathcal{W}$.
A subspace $\mathcal{W}$ is $\mathcal{F}$-invariant, if $\mathcal{W}$ is $A$-invariant, $\forall A \in \mathcal{F}$.
Lemma: If $\mathcal{F}$ is a commuting family, then $\exists \mathbf{x} \in \mathbb{C}^{n}$ that is a common eigenvector $\forall A \in \mathcal{F}$.
Theorem: If $\mathcal{F}$ be a family of diagonalizable matrices, then $\mathcal{F}$ is a commuting family iff it is simultaneously diagonalizable.
Note: Ground field (say $\mathbb{R}$ and $\mathbb{C}$) matters.
Fact: A real symmetric matrix is diagonalizable.
A matrix is (lefty) Markov, if it is nonnegative and its column sums are 1.
Theorem: If $A$ is Markov, then TFAE (the followings are equivalent):