A vector space is a set $V$ along with an addition on $V$ and a scalar multiplication on $V$ such that the following properties hold:

• Commutativity: $u+v=v+u$ for all $u,v\in V$.
• Associativity: $(u+v)+w=u+(v+w)$ and $(ab)v=a(bv)$ for all $u,v,w\in V$ and all $a,b\in \mathbb{F}$.
• Additive Identity: There exists an element $0\in V$ such that $v+0=v$ for all $v\in V$.
• Additive Inverse: For every $v\in V$ there exists $w\in V$ such that $v+w=0$.
• Multiplicative Identity: $1v=v$ for all $v\in V$.
• Distributive Properties: $a(u+v)=au+av$ and $(a+b)v=av+bv$ for all $a,b\in \mathbb{F}$ and all $u,v\in V$.

A subset $U\subseteq V$ is a subspace of $V$ iff $U$ satisfies the following three conditions:

1. Additive Identity: $\overrightarrow{0}\in U$
2. Closed Under Addition: $\overrightarrow{u},\overrightarrow{w}\in U$ implies that $\overrightarrow{u}+\overrightarrow{w}\in U$
3. Closed Under Scalar Multiplication: $a\in \mathbb{F}$ and $\overrightarrow{u}\in U$ implies that $a\overrightarrow{u}\in U$

Suppose $U_1,...,U_m$ are subsets of $V$. The sum of $U_1,...,U_m$, denoted $U_1+...+U_m$, is the set of all possible sums of elements of $U_1,...,U_m$. More precisely:
$U_1+\cdots +U_m=\{u_1+\cdots +u_m:u_1\in U_1,...,u_m\in U_m\}$

Suppose $U_1,\dots ,U_m$ are subspaces of $V$. Then $U_1+\cdots +U_m$ is a direct sum iff the only way to write $0$ as a sum $u_1+\cdots +u_m$, where each $u_j\in U_j$. is by taking each $u_j=0$.

Suppose $U$ and $W$ are subspaces of $V$. Then $U+W$ is a direct sum iff $U\cap W=\{0\}$.

A list $v_1,...,v_m$ of vectors in $V$ is called linearly-independent if the only choice of $a_1,...,a_m\in \mathbb{F}$ that makes $a_1{v}_1+\cdots +a_m{v}_m$ equal 0 is $a_1=\cdots =a_m=0$.
The empty list $()$ is also declared to be linearly independent.

The set of all linear combinations of a list of vectors $v_1,...,v_m$ in $V$ is called the span of $v_1,...,v_m$, denoted $\text{span}(v_1,...,v_m)$. In other words:
$\text{span}(v_1,...,v_m)=\{a_1{v}_1+\cdots +a_m{v}_m:a_1,...,a_m\in \mathbb{F}\}$
The span of the empty list $()$ is defined to be $\{\overrightarrow{0}\}$.

Every LI list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space.

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V=U\oplus W$.

Suppose $V$ is finite-dimensional. Then every LI list of vectors in $V$ with length $\text{dim}(V)$ is a basis of $V$.

Suppose $V$ is finite-dimensional. Then every spanning list of vectors in $V$ with length $\text{dim}(V)$ is a basis of $V$.

A linear map from $V$ to $W$ is a function $T:V\to W$ such that:

1. $T(u+v)=Tu+Tv$ for all $u,v\in V$ (additivity)
2. $T(\lambda v)=\lambda (Tv)$ for all $\lambda \in \mathbb{F},v\in V$. (homogeneity)

Suppose $T$ is a linear map from $V\to W$. Then $T(0)=0$

For $T\in \mathcal{L}(V,W)$ the null space of $T$, denoted $\text{null}(T)$, is the subset of $V$ consisting of those vectors that $T$ maps to ${\overrightarrow{0}}_W$. So $\text{null}(T)=\{v\in V:Tv={\overrightarrow{0}}_W\}$.

Suppose $T\in \mathcal{L}(V,W)$. Then $\text{null}(T)$ is a subspace of $V$.

A function $T:V\to W$ is injective if $Tu=Tv$ implies that $u=v$

Let $T\in \mathcal{L}(V,W)$. Then $T$ is injective iff $\text{null}T=\{\overrightarrow{0}\}$.

For $T:V\to W$ the range of $T$ is the subset of $W$ consisting of those vectors that are of the form $Tv$ for some $v\in V$:
$\text{range}(T)=\{Tv:v\in V\}$

If $T\in \mathcal{L}(V,W)$ then the $\text{range}(T)$ is a subspace of $W$.

A function $T:V\to W$ is surjective if its range equals $W$.

Suppose $V$ is finite-dimensional and $T\in \mathcal{L}(V,W)$. Then $\text{range}(T)$ is finite dimensional and:
$\text{dim}(V)=\text{dim}(\text{null}(T))+\text{dim}(\text{range}(T))$

Suppose $T\in \mathcal{L}(V,W)$ and $v_1,...,v_n$ is a basis of $V$and $w_1,...,w_m$ is a basis of $W$. The matrix of $T$ with respect to these bases is the $m\times n$ matrix $\mathcal{M}(T)$ whose entries $A_{j,k}$ are defined by:
$T{v}_k=A_{1,k}{w}_1+\cdots +A_{m,k}{w}_m$
If the bases are not clear from the context, then the notation $\mathcal{M}(T,(v_1,...,v_n),(w_1,...,w_m))$ is used.

Suppose $A$ is an $m\times n$ matrix and $C$ is $n\times p$. Then $AC$ is defined to be the $m\times p$ matrix whose entry in row $j$ and column $k$ is given by:
$(AC{)}_{j,k}=\sum _{r=1}^n{A}_{j,r}{C}_{r,k}$

Suppose $A$ is $m\times n$. Then:

• If $1\le j\le m$ then $A_{j,\cdot }$ denotes the $1\times n$ matrix consisting of row $j$ of $A$.
• If $1\le k\le n$ then $A_{\cdot ,k}$ denotes the $k\times 1$ matrix consisting of column $k$ of $A$.

• A linear map $T\in \mathcal{L}(V,W)$ is called invertible if there exists a linear map $S\in \mathcal{L}(W,V)$ such that $ST$ equals the identity map on $V$ and $TS$ equals the identity map on $W$.
• A linear map $S\in \mathcal{L}(W,V)$ satisfying $ST=I_V$ and $TS=I_W$ is called the inverse of $T$.

Two finite-dimensional vector spaces over $\mathbb{F}$ are isomorphic iff they have the same dimension.

Suppose $v\in V$ and $v_1,...,v_n$ is a basis for $V$. The matrix of $v$ with respect to this basis is the $n$-by-1 matrix:
$\mathcal{M}(v)=\left(\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right)$
where $c_1,...,c_n$ are the scalars such that:
$v=c_1{v}_1+\cdots +c_n{v}_n$

Suppose $T\in \mathcal{L}(V,W)$ and $v\in V$. Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. Then:
$\mathcal{M}(Tv)=\mathcal{M}(T)\mathcal{M}(v)$

A linear map from a vector space to itself is called an operator. The notation $\mathcal{L}(V)$ denotes the set of all operators on $V$. In other words, $\mathcal{L}(V)=\mathcal{L}(V,V)$.

Suppose $V$ is finite-dimensional and $T\in \mathcal{L}(V)$. Then the following are equivalent:

• $T$ is invertible
• $T$ is injective
• $T$ is surjective

Suppose that $p,s\in \mathcal{P}(\mathbb{F})$ with $s\ne 0$. Then there exist unique polynomials $q,r\in \mathcal{P}(\mathbb{F})$ such that:
$p=sq+r$
where $\text{deg}(r)<\text{deg}(s)$.

Suppose $p\in \mathcal{P}(\mathbb{F})$ and $\lambda \in \mathbb{F}$. Then $p(\lambda )=0$ iff there is a polynomial $q\in \mathcal{P}(\mathbb{F})$ such that:
$p(z)=(z-\lambda )q(z)$
for every $z\in \mathbb{F}$.

A number $\lambda \in \mathbb{F}$ is called a zero (or root) of a polynomial $p\in \mathcal{P}(\mathbb{F})$ if:
$p(\lambda )=0$
On the other hand, a polynomial $s\in \mathcal{P}(\mathbb{F})$ is called a factor of $p$ if there exists a polynomial $q\in \mathcal{P}(\mathbb{F})$ such that $p=sq$.

Suppose $T\in \mathcal{L}(V)$. A subspace $U$ of $V$ is an invariant under $T$ if $u\in U$ implies that $Tu\in U$.

Suppose $T\in \mathcal{L}(V)$. A number $\lambda \in \mathbb{F}$ is called an eigenvalue of $T$ if there exists $v\in V$ such that $v\ne 0$ and $Tv=\lambda v$.

Suppose $V$ is finite-dimensional and $T\in \mathcal{L}(V)$ and $\lambda \in \mathbb{F}$. Then the following are equivalent.

• $\lambda$ is an eigenvalue of $T$;
• $T-\lambda I$ is not injective;
• $T-\lambda I$ is not surjective;
• $T-\lambda I$ is not invertible.

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

Suppose $T\in \mathcal{L}(V)$ and $p\in \mathcal{P}(\mathbb{F})$ is a polynomial given by:
$
p(z) = a_0 + a_1z + a_2z^2 + \dots + a_mz> $
for $z\in \mathbb{F}$. Then $p(T)$ is the operator defined by:
$
p(T) = a_0I + a_1T + a_2T^2 + \dots + a_mT> $

Every operator on a finite-dimensional, nonzero, complex vector space has an eigenvalue.

Suppose $T\in \mathcal{L}(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:

1. The matrix of $T$ ($M(T)$) with respect to $v_1,...,v_n$ is upper-triangular
2. $T{v}_j\in \text{span}(v_1,...,v_j)$ for each $j=1,...,n$
3. $\text{span}(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$

Suppose $T\in \mathcal{L}(V)$ has an upper triangular matrix with respect to some basis of $V$. Then $T$ is invertible iff all the entries on the diagonal of that upper-triangular matrix are non-zero.

Suppose $T\in \mathcal{L}(V)$ has an upper-triangular matrix with respect to some basis of $V$. Then the eigenvalues of $T$ are precisely the entries on the diagonal of that upper-triangular matrix.

Suppose $T\in \mathcal{L}(V)$ and $\lambda \in \mathbb{F}$. The eigenspace of $T$ corresponding to $\lambda$, denoted $E(\lambda ,T)$, is defined by:
$E(\lambda ,T)=\text{null}(T-\lambda I)$
In other words, $E(\lambda ,T)$ is the set of all eigenvectors of $T$ corresponding to $\lambda$, along with the $\overrightarrow{0}$ vector.

Suppose $V$ is finite-dimensional and $T\in \mathcal{L}(V)$. Suppose also that ${\lambda }_1,...,{\lambda }_m$ are distinct eigenvalues of $T$. Then:
$E({\lambda }_1,T)+\cdots +E({\lambda }_m,T)$
is a direct sum, and furthermore,
$\text{dim}(E({\lambda }_1,T))+\cdots +\text{dim}(E({\lambda }_m,T))\le \text{dim}(V)$

If $T\in \mathcal{L}(V)$ has $\text{dim}(V)$ distinct eigenvalues, then $T$ is diagonalizable.

An inner product on $V$ is a function that takes each ordered pair $(u,v)$ of elements of $V$ to a number $⟨u,v⟩\in \mathbb{F}$ and has the following properties:

• (positivity): $⟨v,v⟩\ge 0$ for all $v\in V$
• (definiteness): $⟨v,v⟩=0$ iff $v=0$
• (additivity in the first slot): $⟨u+v,w⟩=⟨u,w⟩+⟨v,w⟩$ for all $u,v,w\in V$
• (homogeneity in the first slot): $⟨\lambda u,v⟩=\lambda ⟨u,v⟩$ for all $\lambda \in \mathbb{F}$ and all $u,v\in V$
• (conjugate symmetry): $⟨u,v⟩=\overline{⟨v,u⟩}$ for all $u,v\in V$

• (a) For each fixed $u\in V$ the function that takes $v$ to $⟨v,u⟩$ is linear from $V$ to $\mathbb{F}$
• (b) $⟨0,u⟩=0$ for every $u\in V$
• (c) $⟨u,0⟩=0$ for every $u\in V$
• (d) $⟨u,v+w⟩=⟨u,v⟩+⟨u,w⟩$ for all $u,v,w\in V$
• (e) $⟨u,\lambda v⟩=\bar{\lambda }⟨u,v⟩$ for all $\lambda \in \mathbb{F}$ and $u,v\in V$

For $v\in V$ the norm of $v$, denoted $||v||$, is defined by:
$||v||=\sqrt{⟨v,v⟩}$

Suppose $v\in V$:

• (a) $||v||=0$ iff $v=0$
• (b) $||\lambda v||=|\lambda |||v||$ for all $\lambda \in \mathbb{F}$

Two vectors $u,v\in V$ are called orthogonal if $⟨u,v⟩=0$.

Suppose $u,v$ are orthogonal vectors in $V$. Then:
$
||u + v||^2 = ||u||^2 + ||v||> $

Suppose $u,v\in V$. Then:
$|⟨u,v⟩|\le ||u||||v||$
This inequality is an equality iff one of $u,v$ is a scalar multiple of the other.

Suppose $u,v\in V$. Then:
$\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$
where we get equality iff one of $u,v$ is a non-negative multiple of the other.

Suppose $u,v\in V$. Then:
$\Vert u+v{\Vert }^2+\Vert u-v{\Vert }^2=2(\Vert u{\Vert }^2+\Vert v{\Vert }^2)$

If $e_1,...,e_m$ is an orthonormal list of vectors in $V$, then:
$
\Vert a_1e_1 + \dots + a_me_m \Vert^2 = |a_1|^2 + \dots + |a_m|> $
for all $a_i\in \mathbb{F}$.

Suppose $e_1,...,e_n$ is an orthonormal basis of $V$ and $v\in V$. Then:
$v=⟨v,e_1⟩e_1+\cdots +⟨v,en⟩e_n$
and:
$
\Vert v \Vert^2 = |\langle v, e_1 \rangle|^2 + \dots + |\langle v, e_n \rangle|> $

Suppose $v_1,...,v_m$ is a LI list of vectors from $V$. Let $e_1=v_1/\Vert v_1\Vert$. For $j=2,...,m$, define $e_j$ inductively by:
$e_j=\frac{v_j-⟨v_j,e_1⟩e_1-\cdots -⟨v_j,e_{j-1}⟩e_{j-1}}{\Vert v_j-⟨v_j,e_1⟩e_1-\cdots -⟨v_j,e_{j-1}⟩e_{j-1}\Vert }$
Then $e_1,...,e_m$ is an orthonromal list of vectors in $V$ such that:
$\text{span}(v_1,...,v_j)=\text{span}(e_1,...,e_j)$
for $j=1,...,m$

Suppose $V$ is finite-dimensional. Then every orthonormal list of vectors in $V$ can be extended to an orthonormal basis of $V$.

Suppose $T\in \mathcal{L}(V)$. If $T$ has an upper-triangular matrix with respect to some basis of $V$, then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.

Suppose $V$ is a finite-dimensional complex vector space and $T\in \mathcal{L}(V)$. Then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.