Linear Algebra
Vector Space
Definition
Motivation: properties of addition and scalar multiplication in $ \mathbf{F}^n $
flowchart LR
subgraph addGroup["$$u + v$$"]
direction TB
addSet["abelian group under addition"]
end
subgraph scalarGroup["$$\lambda v$$"]
direction TB
scalarSet["• associativity<br/>• 1"]
end
addSet <-->|distributive<br/>properties| scalarSet
style addGroup fill:transparent,stroke:transparent
style scalarGroup fill:transparent,stroke:transparent
Examples
- $ \mathbf{F}^S $
- $ \mathcal{P}(\mathbf{F}) $
- $ \mathcal{P}_m(\mathbf{F}) $
- $ \mathcal{L}(V, W) $ ($ \dim = (\dim V)(\dim W) $)
- $ \mathbf{F}^{m,n} $ ($ \dim = mn $)
- $ V_1 \times \dots \times V_m $ ($ \dim = \dim V_1 + \dots + \dim V_m $)
- $ V/U $ ($ \dim = \dim V - \dim U $)
Subspace
Examples
- $ V_1 \cap \dots \cap V_m $
- $ V_1 \cup V_2 $ ($ \Leftrightarrow V1 \subseteq V2 $ or $ V1 \supseteq V2 $)
- $ V_1 + \dots + V_m $ (Smallest containing subspace)
- $ U^0 $ ($ \dim = \dim V - \dim U $)
Suppose $V$ is finite-dimensional. $ U $ is a subspace of $ V $:
- $ \dim U \leq \dim V $
- $ \dim U = \dim V \Leftrightarrow U = V $
Sum
Analogy between sets and vector spaces:
- Cardinality: Dimension
- Union: Sum
- Union of disjoint sets: Direct sum
Number of Vectors
The number of vectors equals the number of choices of the coefficient tuple $(a_1, \dots, a_n)$:
\[\#(\text{vectors}) = \lvert \mathbf{F} \rvert ^n\]Finite fields exist precisely for prime-power sizes $q = p^k$.
| over $\mathbb{R}$ or $\mathbb{C}$ | over $\mathbf{F}_q$ | |
|---|---|---|
| trivial space | $1$ | $1$ |
| dimension $n \geq 1$ | $\infty$ | $q^n$ |
Bases
A list is a basis if it satisfies any two of the following three conditions:
- It is linearly independent
- It spans $V$
- Its length equals $ \dim(V) $
A direct-sum decomposition of $ V $ is the same thing as a partition of a basis of $ V $.
A sum is a direct sum iff dimensions add up.
Linear Maps
Examples
- $ T : V \to W $
- $ \Gamma : V_1 \times \dots \times V_m \to V_1 + \dots + V_m $ (Invertible iff $ V_1, \dots, V_m $ is a direct sum)
- Quotient map: $ \pi : V \to V/U $, $ \pi(v) = v + U $
- $ \tilde{T} ∶ V/(\operatorname{null} T) \to W $, $ \tilde{T}(v + \operatorname{null} T) = Tv $
- $ \tilde{T} \circ \pi = T $
- $ T \to T’ $
- $ P_U \in \mathcal{L}(V) $
- $ (\operatorname{null} T)^{\perp} \to \operatorname{range} T $ (Invertable)
- $ T^{\ast} $
A linear map may be prescribed freely on a basis:
$ \forall \ \text{basis} \ v_1, \dots, v_n \in V $ and $ \forall w_1, \dots, w_n \in W $, $ \exists! T \in \mathcal{L}(V, W) \ \text{s.t.} $
\[Tv_k = w_k\]for each $k = 1, \dots, n$
Extension
$ V $ is finite-dimensional:
$ U $ is a subspace of $ V $, $ S \in \mathcal{L}(U, W) \Rightarrow \exists T \in \mathcal{L}(V, W) \ \text{s.t.} \ T \vert _U = S $
Extend a basis $ u_1,\dots,u_m$ of $U $ to a basis $ u_1,\dots,u_m,\,v_1,\dots,v_n $ of $ V $. Define $ T u_i = S u_i $ and $ T v_j = $ anything in $ W $; extend linearly.
Algebraic Operations
- $ \mathcal{L}(V, W) $ is a vector space
- Product of linear maps is a bilinear map
Null Spaces and Ranges
Fundamental theorem of linear maps
Suppose $V$ is finite-dimensional. $ T \in \mathcal{L}(V, W) $:
\[\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T\]
Generalization
Suppose $V$ is finite-dimensional. $ T \in \mathcal{L}(V, W) $, $ U $ is a subspace of $ W $:
\[\dim \{ v \in V : Tv \in U \} = \dim \operatorname{null} T + \dim (U \cap \operatorname{range} T)\]
Injectivity, Surjectivity and Invertibility
Suppose $ T \in \mathcal{L}(V, W) $.
(a) If $ T $ is injective, then it’s invertible from $ V $ to $ \operatorname{range} T $.
(b) Decompose $ V = \operatorname{null} T \oplus U $, then $(\left. T \right\rvert_{U})$ is invertible.
$ \Leftrightarrow $
| $ T \in \mathcal{L}(V, W) $ | Injective | Surjective | Invertible (Isomorphic) |
|---|---|---|---|
| Definition | $ \operatorname{null} T = {0} $ | $ \operatorname{range} T = W $ | $ T $ is injective and $ T $ is surjective |
| Preservation | linear independence | spanning | basis |
| Inverse | $ T $ has a left inverse: $ ST = I $ | $ T $ has a right inverse: $ TS = I $ | $ T $ has the inverse: $ ST = I $ and $ TS = I $ |
$ \Rightarrow $
| $ T \in \mathcal{L}(V, W) $ | Injective | Surjective | Invertible (Isomorphic) |
|---|---|---|---|
| Dimensions (Finite) | $ \dim V \leq \dim W $ | $ \dim V \geq \dim W $ | $ \dim V = \dim W $ |
| $\dim V = \dim W$ | (all three coincide) | (all three coincide) | (all three coincide) |
| Pseudoinverse | $ TT^{\dagger} = I $ | $ T^{\dagger}T = I $ | $ TT^{\dagger} = T^{\dagger}T = I $ |
Product
Suppose $U$ and $V$ are finite-dimensional. $ S \in \mathcal{L}(V, W) $ and $ T \in \mathcal{L}(U, V) $:
- $ \operatorname{null} T \subseteq \operatorname{null} ST $
- $ \operatorname{range} ST \subseteq \operatorname{range} S $
Suppose $ U $ and $ V $ are finite-dimensional. $ S \in \mathcal{L}(V, W) $ and $ T \in \mathcal{L}(U, V) $:
- $ \dim \operatorname{null} ST \leq \dim \operatorname{null} S + \dim \operatorname{null} T $
- $ \dim \operatorname{range} ST \leq \min(\dim \operatorname{range} S + \dim \operatorname{range} T) $
Suppose $ V $ is finite-dimensional and $ S, T \in \mathcal{L}(V, W) $:
- $ ST $ is invertible $\Leftrightarrow$ $ S $ and $ T $ are invertible.
Translate
Re-basing
Suppose $ U $ is a subspace of $ V $ and $ v,w \in V $. Then
\[x \in v + U \Leftrightarrow x + U = v + U.\]
Corollary: Two translates of a subspace are equal or disjoint.
\[v - w \in U \Leftrightarrow v + U = w + U \Leftrightarrow (v + U) \cap (w + U) \ne \emptyset\] \[v \in U \Leftrightarrow v + U = 0 + U.\]
Suppose $ A_1 = v + U_1 $ and $ A_2 = w + U_2 $ for some $ v,w \in V $ and some subspaces $ U_1,U_2 $ of $ V $.
\[\forall x \in A_1 \cap A_2, \ A_1 \cap A_2 = (x + U_1) \cap (x + U_2) = x + (U_1 \cap U_2).\]
Suppose $ T \in \mathcal{L}(V,W) $ and $ c \in W $:
$ {x \in V : Tx = c} $ is either the empty set or is a translate of $ \operatorname{null} T $.
Special case: system of linear equations
general solution = particular solution + homogeneous solution
$ V/\operatorname{null} T \cong_\tilde{T} \operatorname{range} T $
Bases
Suppose $ U $ and $ W $ are subspaces of $ V $ and $ V = U \oplus W $. $\pi|_W : W \to V/U$ is an isomorphism.
Duality (Project, function)
Suppose $ U $ and $ W $ are subspaces of $ V $ and $ V = U \oplus W $. Suppose $ w_1, \dots, w_m $ is a basis of $ W $. Then $ w_1 + U, \dots, w_m + U $ is a basis of $ V/U $.
Duality (Lift, one-to-many)
Suppose that $ U $ is a subspace of $ V $ such that $ V/U $ is finite-dimensional. There exists a finite-dimensional subspace $ W $ of $ V $ such that $ \dim W = \dim V/U $ and $ V = U \oplus W $.
Since $V/U$ is finite-dimensional, pick a basis
\[v_1 + U,\ \dots,\ v_m + U \quad (m = \dim V/U),\]with chosen representatives $v_1, \dots, v_m \in V$. Define
\[W := \operatorname{span}(v_1, \dots, v_m).\]We show this $W$ works: it’s finite-dimensional, $\dim W = m = \dim V/U$, and $V = U \oplus W$.
The $v_k$ are linearly independent (so $\dim W = m$). Suppose $\sum_k a_k v_k = 0$. Apply $\pi$:
\[0 + U = \pi\Big(\sum_k a_k v_k\Big) = \sum_k a_k (v_k + U).\]Since $v_1 + U, \dots, v_m + U$ is a basis of $V/U$, it’s independent, so all $a_k = 0$. Thus $v_1, \dots, v_m$ is independent and $\dim W = m = \dim V/U$.
$U + W = V$. Let $v \in V$. Expand its coset in the basis: $v + U = \sum_k a_k (v_k + U) = \big(\sum_k a_k v_k\big) + U$. Equal cosets differ by an element of $U$:
\[v - \sum_k a_k v_k \in U \quad\Longrightarrow\quad v = \underbrace{\Big(v - \sum_k a_k v_k\Big)}_{\in\, U} + \underbrace{\sum_k a_k v_k}_{\in\, W} \in U + W.\]$U \cap W = {0}$. Let $x \in U \cap W$. Since $x \in W$, write $x = \sum_k a_k v_k$. Since $x \in U$, $\pi(x) = 0$:
\[0 + U = \pi(x) = \sum_k a_k (v_k + U).\]Independence of the basis forces all $a_k = 0$, so $x = 0$.
Therefore $V = U \oplus W$ with $\dim W = \dim V/U$. $\blacksquare$
Suppose $ U $ is a subspace of $ V $ and $ v_1 + U, \dots, v_m + U $ is a basis of $ V/U $ and $ u_1, \dots, u_n $ is a basis of $ U $. Then $ v_1, \dots, v_m, u_1, \dots, u_n $ is a basis of 𝑉.
Isomorphism
- $ V/\operatorname{null} T \cong_\tilde{T} \operatorname{range} T $
- $ \mathcal{L}(V, W) \cong \mathbf{F}^{m,n} $
- $ V \cong \mathcal{L}(\mathbf{F}, V) $
- $ V^m \cong \mathcal{L}(\mathbf{F}^m, V) $
- $ V \cong U \times V/U $
Matrices
$ \mathcal{M}(T,(v_1,\dots,v_n),(w_1,\dots,w_m)) \in \mathbf{F}^{m,n} $ is the matrix representation of $ T $ with respect to the chosen bases.
\[Tv_k = \sum_{j=1}^m {A_{j,k}w_j}\]$ T \in \mathcal{L}(\mathbf{F}^n, \mathbf{F}^m) $, $ \mathcal{M}(T,(e_1, \dots, e_n), (e_1, \dots, e_m))_{\cdot,k} = Te_k $
Column of matrix product equals matrix times column
Linear combination of columns
$ \dim \operatorname{range} S = \text{rank} \, \mathcal{M}(T) $
Column–row factorization
$ C $ is a basis of the column space.
$ A \mapsto A^T $ is a linear map
\[\mathcal{M}(I,(u_1,\dots,u_n),(v_1,\dots,v_n))\mathcal{M}(I,(v_1,\dots,v_n),(u_1,\dots,u_n)) = I\]Change-of-basis
$ T \in \mathcal{L}(V) $, $ A = \mathcal{M}(T,(u_1,\dots,u_n)), B = \mathcal{M}(T,(v_1,\dots,v_n)), C = \mathcal{M}(I,(u_1,\dots,u_n),(v_1,\dots,v_n))$:
\[A = C^{-1}BC\]Dual Space and Dual Map
- $ T $ is injective $ \Leftrightarrow $ $ T’ $ is surjective
$ T $ is surjective $ \Leftrightarrow $ $ T’ $ is injective
- $ \operatorname{null} T’ = (\operatorname{range} T)^0 $
$ \operatorname{range} T’ = (\operatorname{null} T)^0 $
- $ \dim \operatorname{null} T’ = \dim \operatorname{null} T + \dim W - \dim V $
- $ \dim \operatorname{range} T’ = \dim \operatorname{range} T $





