Suppose $V_1,...,V_m$ are vector spaces over $\mathbb{F}$.

• The product $V_1\times \cdots \times V_m$ is defined by:
  $V_1\times \cdots \times V_m=\{(v_1,...,v_m):\mathrm{\forall }i(v_i\in V_i)\}$
• Addition on $V_1\times \cdots \times V_m$ is defined by:
  $(u_1,...,u_m)+(v_1,...,v_m)=(u_1+v_1,...,u_m+v_m)$
• Scalar multiplication on $V_1\times \cdots \times V_m$ is defined by:
  $\lambda (v_1,...,v_m)=(\lambda v_1,...,\lambda v_m)$

Suppose $V_1,...,V_m$ are vector spaces over $\mathbb{F}$. Then $V_1\times \cdots \times V_m$ is a vector space over $\mathbb{F}$.

Suppose $V_1,...,V_m$ are finite-dimensional vector spaces. Then $V_1\times \cdots \times V_m$ is finite-dimensional and:
$\text{dim}(V_1\times \cdots \times V_m)=\text{dim}{V}_1+\cdots +\text{dim}{V}_m$

Suppose that $U_1,...,U_m$ are subspaces of $V$. Define a linear map $\mathrm{\Gamma }:U_1\times \cdots \times U_m\to U_1+\cdots +U_m$ by:
$\mathrm{\Gamma }(u_1,...,u_m)=u_1+\cdots +u_m$
Then $U_1+\cdots +U_m$ is a direct sum iff $\mathrm{\Gamma }$ is injective/invertible

Suppose $V$ is finite-dimensional and $U_1,...,U_m$ are subspaces of $V$. Then $U_1+\cdots +U_m$ is a direct sum iff:
$\text{dim}(U_1+\cdots +U_m)=\text{dim}{U}_1+\cdots +\text{dim}{U}_m$

Suppose $v\in V$ and $U$ is a subspace of $V$. Then $v+U$ is the subset of $V$ defined by:
$v+U=\{v+u:u\in U\}$

• An affine subset of $V$ is a subset of $V$ of the form $v+U$ for some $v\in V$ and some subspace $U$ of $V$
• For $v\in V$ and $U$ being a subspace of $V$, the affine subset $v+U$ is said to be parallel to $U$.

Suppose $U$ is a subspace of $V$. Then the quotient space $V/U$ is the set of all affine subsets of $V$ parallel to $U$. In other words:
$V/U=\{v+U:v\in V\}$

Suppose $U$ is a subspace of $V$ and $v,w\in V$. Then the following are equivalent:

• $v-w\in U$
• $v+U=w+U$
• $(v+U)\cap (w+U)\ne \mathrm{\varnothing }$

Suppose $U$ is a subspace of $V$. Then addition and scalar multiplication are defined on $V/U$ by:
$\begin{array}{rl}(v+U)+(w+U)& =(v+w)+U\\ \lambda (v+U)& =(\lambda v)+U\end{array}$
for $v,w\in V$ and $\lambda \in \mathbb{F}$.

Suppose $U$ is a subspace of $V$. Then $V/U$, with the operations of addition and scalar multiplication as defined above, is a vector space.

Suppose $U$ is a subspace of $V$. The quotient map $\pi$ is the linear map $\pi :V\to V/U$ defined by:
$\pi (v)=v+U$
for $v\in V$.

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then:
$\text{dim}(V/U)=\text{dim}(V)-\text{dim}(U)$

Suppose $T\in \mathcal{L}(V,W)$. Define $\tilde{T}:V/(\text{null}(T))\to W$ by:
$\tilde{T}(v+\text{null}(T))=Tv$

Suppose $T\in \mathcal{L}(V,W)$. Then:

• $\tilde{T}$ is a linear map from $V/(\text{null}(T))$ to $W$
• $\tilde{T}$ is injective
• $\text{range}(\tilde{T})=\text{range}(T)$
• $V/(\text{null}(T))$ is isomorphic to $\text{range}(T)$.

A linear functional of vector space $V$ is an element of $\mathcal{L}(V,\mathbb{F})$.

The dual space of $V$, denoted $V^{\prime }$, is the vector space of all linear functionals on $V$. In other words, $V^{\prime }=\mathcal{L}(V,\mathbb{F})$

Suppose $V$ is finite-dimensional. Then $V^{\prime }$ is also finite dimensional and is the same dimension as $V$.

If $v_1,...,v_n$ is a basis of $V$ then the dual basis of $v_1,...,v_n$ is the list ${\phi }_1,...,{\phi }_n$ of elements of $V^{\prime }$ where each ${\varphi }_j$ is the linear functional on $V$ such that:
${\varphi }_j(v_k)=\{\begin{array}{ll}1& k=j\\ 0& k\ne j\end{array}=\chi (k=j)$

Suppose $V$ is finite-dimensional. Then the dual basis of a basis of $V$ is a basis of $V^{\prime }$.

If $T\in \mathcal{L}(V,W)$ then the dual map of $T$ is the linear map $T^{\prime }\in \mathcal{L}(W^{\prime },V^{\prime })$ defined by $T^{\prime }(\phi )=\phi \circ T$ for $\phi \in W^{\prime }$.

• $(S+T{)}^{\prime }=S^{\prime }+T^{\prime }$ for all $S,T\in \mathcal{L}(V,W)$
• $(\lambda T{)}^{\prime }=\lambda T^{\prime }$ for all $\lambda \in \mathbb{F},T\in \mathcal{L}(V,W)$
• $(ST{)}^{\prime }=T^{\prime }{S}^{\prime }$ for all $T\in \mathcal{L}(U,V)$ and $S\in \mathcal{L}(V,W)$

For $U\subset V$ the annihilator of $U$, denoted $U^0$, is defined by:
$U^0=\{\phi \in V^{\prime }:\phi (u)=0\mathrm{\forall }u\in U\}$

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then:
$\text{dim}(U)+\text{dim}(U^0)=\text{dim}(V)$

Suppose $V,W$ are finite dimensional and $T\in \mathcal{L}(V,W)$. Then:

• $\text{dim}(\text{range}(T^{\prime }))=\text{dim}(\text{range}(T))$
• $\text{range}(T^{\prime })=(\text{null}(T){)}^0$

Suppose $V,W$ are finite-dimensional and $T\in \mathcal{L}(V,W)$. Then $T$ is surjective iff $T^{\prime }$ is injective.

Suppose $V,W$ are finite-dimensional and $T\in \mathcal{L}(V,W)$. Then $T$ is injective iff $T^{\prime }$ is surjective.

Suppose $T\in \mathcal{L}(V,W)$. Then $\mathcal{M}(T^{\prime })=(\mathcal{M}(T){)}^T$.

The rank of a matrix $A\in {\mathbb{F}}^{m,n}$ is the column rank of $A$.

Suppose $A$ is an $m\times n$ matrix with entries in $\mathbb{F}$.

• The row rank of $A$ is the dimension of the span of the rows of $A$ in ${\mathbb{F}}^{1,n}$.
• The column rank of $A$ is the dimension of the span of the columns of $A$ in ${\mathbb{F}}^{m,1}$.

Suppose $V$ is finite-dimensional and $\phi$ is a linear functional on $V$. Then there is a unique vector $u\in V$ such that:
$\phi (v)=⟨v,u⟩$
for every $v\in V$.

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 $v\in V,w\in W$.

If $T\in \mathcal{L}(V,W)$ then $T^{\ast }\in \mathcal{L}(W,V)$.

For all $S,T\in \mathcal{L}(V,W)$ and $\lambda \in \mathbb{F}$:

• $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$
• $(\lambda T{)}^{\ast }=\bar{\lambda }{T}^{\ast }$
• $(T^{\ast }{)}^{\ast }=T$
• $I^{\ast }=I$ where $I$ is the identity operator on $V$.
• If instead $S\in \mathcal{L}(W,U)$, then $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Here $U$ is an inner product space over $\mathbb{F}$.

Suppose $T\in \mathcal{L}(V,W)$. Then:

• $\text{null}(T^{\ast })=(\text{range}(T){)}^{\perp }$
• $\text{range}(T^{\ast })=(\text{null}(T){)}^{\perp }$
• $\text{null}(T)=(\text{range}(T^{\ast }){)}^{\perp }$
• $\text{range}(T)=(\text{null}(T^{\ast }){)}^{\perp }$

The conjugate transpose of an $m\times n$ matrix is the $n\times m$ matrix obtained by interchanging the rows and columns and then taking the complex conjugate of each entry.

Let $T\in \mathcal{L}(V,W)$. Suppose $e_1,...,e_n$ is an orthonormal basis of $V$ and $f_1,...,f_m$ is an orthonormal basis of $W$. Then:
$\mathcal{M}(T^{\ast },(f_1,...,f_m),(e_1,...,e_m))$
is the conjugate transpose of:
$\mathcal{M}(T,(e_1,...,e_n),(f_1,...,f_m))$

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 $v,w\in V$

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 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 normal iff:
$\Vert Tv\Vert =\Vert T^{\ast }v\Vert$
for all $v\in V$.

Suppose $T\in \mathcal{L}(V)$ is normal and $v\in V$ is an eigenvector of $T$ with eigenvalue $\lambda$. Then $v$ is also an eigenvector of $T^{\ast }$ with eigenvalue $\bar{\lambda }$.

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

Suppose $T\in \mathcal{L}(V)$ is normal. Then:
$\text{range}(T)=\text{range}(T^{\ast })$

Suppose $T\in \mathcal{L}(V)$ is normal. Then:
$\text{null}(T^k)=\text{null}(T),\text{range}(T^k)=\text{range}(T)$
for all $k\in {\mathbb{N}}^{+}$.

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 $V$

Suppose $T\in \mathcal{L}(V)$ is self-adjoint and $U$ is a subspace of $V$ that is invariant under $T$. Then:

• $U^{\perp }$ is invariant under $T$.
• $T{|}_U\in \mathcal{L}(U)$ is self-adjoint
• $T{|}_{U^{\perp }}\in \mathcal{L}(U^{\perp })$ is self-adjoint.

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 $R^2=T$.

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. $T=R^{\ast }R$.

Every positive operator on $V$ has a unique positive square root.

• 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.

• $S$ 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 $V$ is a complex inner product space and $S\in \mathcal{L}(V)$. Then the following are equivalent:

• $S$ is an isometry
• There is an orthonormal basis of $V$ consisting of eigenvectors of $S$ whose corresponding eigenvalues all have absolute value of 1.

If $T$ is a positive operator then $\sqrt{T}$ denotes the unique positive square root of $T$.

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

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

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 $G(\lambda ,T)=\text{null}(T-\lambda I{)}^{\dim (V)}$.

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.

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

Suppose $N\in \mathcal{L}(V)$ is nilpotent. Then $N^{\dim (V)}=\overrightarrow{0}$

Suppose $N$ is a nilpotent operator on $V$. Then $\mathrm{\exists }$ a basis of $V$ w.r.t. which the matrix of $N$ has the form:
$\left[\begin{array}{ccc}0& & \ast \\ & \ddots & \\ 0& & 0\end{array}\right]$
so all entries on and below the diagonal are 0's.

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.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Then there is a basis of $V$ consisting of generalized eigenvectors of $T$.

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 $\dim (\text{null}(T-\lambda I{)}^{\dim (V)})$.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Then the sum of the multiplicities of all eigenvalues of $T$ equals $\dim (V)$.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Let ${\lambda }_1,...,{\lambda }_m$ be the distinct eigenvalues of $T$, with multiplicities $d_1,...,d_m$. Then there is a basis of $V$ with respect to which $T$ has a blcok diagonal matrix like seen above:
$\left[\begin{array}{ccc}{A}_1& \dots & 0\\ ⋮& \ddots & ⋮& \\ 0& \dots & A_m\end{array}\right]$
where each $A_j$ is a $d_j\times d_j$ upper-triangular matrix:
$A_j=\left[\begin{array}{ccc}{\lambda }_j& \dots & \ast \\ ⋮& \ddots & ⋮\\ 0& \dots & {\lambda }_j\end{array}\right]$

Suppose $N\in \mathcal{L}(V)$ is nilpotent. Then $I+N$ has a square root.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$ is invertible. Then $T$ has a square root.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Let ${\lambda }_1,...,{\lambda }_m$ denote the distinct eigenvalues of $T$ with multiplicities $d_1,...,d_m$. The polynomial:
$(z-{\lambda }_1{)}^{d_1}\dots (z-{\lambda }_m{)}^{d_m}$
is called the characteristic polynomial of $T$.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Then:

• the characteristic polynomial of $T$ has degree $\dim (V)$
• the zeroes of the characteristic polynomial of $T$ are the eigenvalues of $T$.

Suppose $V$ is a complex vector space and $T\in \mathcal{L}(V)$. Let $q$ denote the characteristic polynomial of $T$. Then $q(T)=0$.

Suppose $T\in \mathcal{L}(V)$. Then there is a unique monic polynomial $p$ of smallest degree such that $p(T)=\overrightarrow{0}$.

1. $\mathrm{deg}({\mu }_T)\le \mathrm{deg}(p_T)=\dim (V)$
2. Any polynomial $s$ with $s(T)={\overrightarrow{0}}_T$ can only happen iff $s$ is a multiple of ${\mu }_T$.
3. $p_T$ is a multiple of ${\mu }_T$
4. The zeroes of ${\mu }_T$ are precisely the eigenvalues of $T$.

Suppose $N\in \mathcal{L}(V)$ is nilpotent. Then $\mathrm{\exists }{v}_1,...,v_n\in V$ and $m_1,...,m_n\in {\mathbb{Z}}^{\ge 0}$ such that:

1. $N^{m_1}{v}_1,...,N{v}_1,v_1,...,N^{m_n}{v}_n,...,N{v}_n,v_n$ is a basis of $V$
2. $N^{m_1+1}{v}_1=\cdots =N^{m_n+1}{v}_n=0$

Suppose $T\in \mathcal{L}(V)$. A basis of $V$ is called a Jordan basis for $T$ if w.r.t. this basis then $T$ has a block diagonal matrix:
$\left[\begin{array}{ccc}{A}_1& & 0\\ & \ddots & \\ 0& & A_p\end{array}\right]$
where each $A_j$ is an UT matrix of the form:
$A_j=\left[\begin{array}{cccc}{\lambda }_j& 1& & 0\\ & \ddots & \ddots & \\ & & \ddots & 1\\ 0& & & {\lambda }_j\end{array}\right]$

1. Choose $v_1\in V$ such that $T^{n-1}v\ne \overrightarrow{0}$ for $n=\dim (V)$.
2. Find a basis for $\text{null}(T^{n-k})/\text{null}(T^{n-k-1})$ where $k\ge 1$.
3. Repeat for growing values of $k$, moving over the difference in dimensions of our nullspace chain.

Suppose $V$ is a real vector space.

• The complexification of $V$, denoted $V_C$, equals $V\times V$. An element of $V_C$ is an ordered pair $(u,v)$ where $u,v\in V$, but we'll write this as $u+iv$.
• Addition on $V_C$ is defined by:
  $(u_1+i{v}_1)+(u_2+i{v}_2)=(u_1+u_2)+i(v_1+v_2)$
  for $u_1,v_1,u_2,v_2\in V$.
• Complex scalar multiplication on $V_C$ is defined by:
  $(a+bi)(u+iv)=(au-bv)+i(av+bu)$
  for $a,b\in \mathbb{R}$ and $u,v\in V$.

Suppose $V$ is a real vector space and $T\in \mathcal{L}(V)$. The complexification of $T$, denoted $T_C$, is the operator $T_C\in \mathcal{L}(V_C)$ defined by:
$T_C(u+iv)=Tu+iTv$
for $u,v\in V$.

Suppose $V$ is a real vector space and $T\in \mathcal{L}(V)$. Then the characteristic polynomial of $T$ is defined by the characteristic polynomial of $T_C$.

Suppose $V$ is a real vector space and $T\in \mathcal{L}(V)$. Then:

• the coefficients of the characteristic polynomial of $T$ are all real.
• the characteristic polynomial of $T$ has degree $\dim (V)$
• The eigenvalues of $T$ are precisely the real zeros of the characteristic polynomial of $T$.

Suppose $T\in \mathcal{L}(V)$. Then:

• The degree of the minimal polynomial of $T$ is at most $\dim (V)$
• The characteristic polynomial of $T$ is a polynomial multiple of the minimal polynomial of $T$.