Suppose $T\in \mathcal{L}(V,W)$. The adjoint of $T$ is the function $T^{\ast }:W\to V$ such that:
$⟨Tv,w⟩=⟨v,T^{\ast }w⟩$
for all $v\in V,w\in W$.

An operator $T\in \mathcal{L}(V)$ is called self-adjoint if $T=T^{\ast }$. In other words, $T\in \mathcal{L}(V)$ is self-adjoint iff:
$⟨Tv,w⟩=⟨v,Tw⟩$
for all $v,w\in V$

• An operator on an inner product space is called normal if it commutes with its adjoint.
• $T\in \mathcal{L}(V)$ is normal if $T{T}^{\ast }=T^{\ast }T$.

An operator $T\in \mathcal{L}(V)$ is called positive if $T$ is self-adjoint and:
$⟨Tv,v⟩\ge 0$
for all $v\in V$.

An operator $R$ is called a square root of an operator $T$ if $R^2=T$.

• An operator $S\in \mathcal{L}(V)$ is called an isometry if:
  $\Vert Sv\Vert =\Vert v\Vert$
  for all $v\in V$.
• In other words, an operator is an isometry if it preserves norms.

Suppose $T\in \mathcal{L}(V)$. Then $\mathrm{\exists }$ an isometry $S\in \mathcal{L}(V)$ such that:
$T=S\sqrt{T^{\ast }T}$

Suppose $T\in \mathcal{L}(V)$. The singular values of $T$ are the eigenvalues of $\sqrt{T^{\ast }T}$ with each eigenvalues $\lambda$ repeated $\text{dim}(E(\lambda ,\sqrt{T^{\ast }T}))$ time.

Suppose $T\in \mathcal{L}(V)$ and $\lambda$ is an eigenvalue of $T$. A vector $v\in V$ is called a generalized eigenvector of $T$ corresponding to $\lambda$ if $v\ne \overrightarrow{0}$ and:
$(T-\lambda I{)}^{j}v=\overrightarrow{0}$
for some $j\in {\mathbb{Z}}^{+}$.

Suppose $T\in \mathcal{L}(V)$ and $\lambda \in \mathbb{F}$. The generalized eigenspace of $T$ corresponding to $\lambda$, denoted $G(\lambda ,T)$, is defined to be the set of all generalized eigenvectors of $T$ corresponding to $\lambda$, along with the $\overrightarrow{0}$.

Suppose $T\in \mathcal{L}(V)$ and $\lambda \in \mathbb{F}$. Then $G(\lambda ,T)=\text{null}(T-\lambda I{)}^{\dim (V)}$.

An operator is called nilpotent if some power of it equals $\overrightarrow{0}$.

Suppose $T\in \mathcal{L}(V)$. The multiplicity of an eigenvalue $\lambda$ of $T$ is defined to be the dimension of the corresponding generalized eigenspace $G(\lambda ,T)$. In other words, the multiplicity of an eigenvalue $\lambda$ of $T$ equals $\dim (\text{null}(T-\lambda I{)}^{\dim (V)})$.

A block diagonal matrix is a square matrix of the form:
$\left[\begin{array}{ccc}{A}_1& \dots & 0\\ ⋮& \ddots & ⋮& \\ 0& \dots & A_m\end{array}\right]$
where $A_1,...,A_m$ are square matrices lying along the diagonal and all the other entries of the matrix equal 0.

Every eigenvalue of a self-adjoint operator is real

Suppose $V$ is a complex inner product space and $T\in \mathcal{L}(V)$. Then $T$ is self-adjoint iff:
$⟨Tv,v⟩\in \mathbb{R}$
for every $v\in V$.

An operator $T\in \mathcal{L}(V)$ is normal iff:
$\Vert Tv\Vert =\Vert T^{\ast }v\Vert$
for all $v\in V$.

Suppose $T\in \mathcal{L}(V)$ is normal. Then eigenvectors of $T$ corresponding to distinct eigenvalues are orthogonal.

Suppose $\mathbb{F}=\mathbb{C}$ and $T\in \mathcal{L}(V)$. Then the following are equivalent:

• $T$ is normal
• $V$ has an orthonormal basis consisting of eigenvectors of $T$.
• $T$ has a diagonal matrix with respect to some orthonormal basis of $V$.

Suppose $\mathbb{F}=\mathbb{R}$ and $T\in \mathcal{L}(V)$. Then the following are equivalent:

• $T$ is self-adjoint
• $V$ has an orthonormal basis consisting of eigenvectors of $T$.
• $T$ has a diagonal matrix with respect to some orthonormal basis of $V$

Let $T\in \mathcal{L}(V)$. Then the following are equivalent:

• $T$ is positive
• $T$ is self-adjoint and all eigenvalues of $T$ are non-negative
• $T$ has a positive square root
• $T$ has a self-adjoint square root
• $\mathrm{\exists }R\in \mathcal{L}(V)$ s.t. $T=R^{\ast }R$.

• $S$ is an isometry
• $⟨Su,Sv⟩=⟨u,v⟩$ for all $u,v\in V$
• $S{e}_1,...,S{e}_n$ is orthonormal for every orthonormal list of vectors $e_1,...,e_n$ in $V$
• $\mathrm{\exists }$ an orthonormal basis $e_1,...,e_n$ of $V$ such that $S{e}_1,...,S{e}_n$ is orthonormal.
• $S^{\ast }S=I$
• $S{S}^{\ast }=I$
• $S^{\ast }$ is an isometry
• $S$ is invertible and $S^{-1}=S^{\ast }$.

Suppose $T\in \mathcal{L}(V)$ has singular values $s_1,...,s_n$. Then there exist orthonormal bases $e_1,...,e_n$ and $f_1,...,f_n$ such that:
$Tv=s_1⟨v,e_1⟩f_1+\cdots +s_n⟨v,e_n⟩f_n$

for every $v\in V$.

Suppose $T\in \mathcal{L}(V)$. Then:
$\{\overrightarrow{0}\}=\text{null}(T^0)\subseteq \text{null}(T^1)\subseteq \cdots \subseteq \text{null}(T^k)\subseteq \text{null}(T^{k+1})\subseteq \dots$

Suppose $T\in \mathcal{L}(V)$. Suppose $m$ is a nonnegative integer such that $\text{null}(T^m)=\text{null}(T^{m+1})$. Then:
$\text{null}(T^m)=\text{null}(T^{m+1})=\dots$

Suppose $T\in \mathcal{L}(V)$. Let $n=\dim (V)$. Then:
$\text{null}(T^n)=\text{null}(T^{n+1})=\dots$

Let $T\in \mathcal{L}(V)$. Suppose ${\lambda }_1,...,{\lambda }_m$ are distinct eigenvalues of $T$ and $v_1,...,v_m$ are corresponding generalized eigenvectors. Then $v_1,...,v_m$ is linearly independent.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Let ${\lambda }_1,...,{\lambda }_m$ be the distinct eigenvalues of $T$.

• $V=\underset{i=1}{\overset{m}{⨁}}G({\lambda }_i,T)$
• Each $G({\lambda }_i,T)$ is invariant under $T$.
• Each $(T-{\lambda }_{i}I){|}_{G({\lambda }_i,T)}$ is nilpotent.