MATH661

1.Partition of linear mapping domain and codomain

A partition of a set $S$ has been introduced as a collection of subsets $P = {S_{i} | S_{i} \subset P, S_{i} \neq \emptyset .}$ such that any given element $x \in S$ belongs to only one set in the partition. This is modified when applied to subspaces of a vector space, and a partition of a set of vectors is understood as a collection of subsets such that any vector except $𝟎$ belongs to only one member of the partition.

Linear mappings between vector spaces $𝒇 : U \to V$ can be represented by matrices $𝑨$ with columns that are images of the columns of a basis ${𝒖_{1}, 𝒖_{2}, \dots}$ of $U$

𝑨 = [\begin{array}{ccc} 𝒇 (𝒖_{1}) & 𝒇 (𝒖_{2}) & \dots \end{array}] .

Consider the case of real finite-dimensional domain and co-domain, $𝒇 : ℝ^{n} \to ℝ^{m}$ , in which case $𝑨 \in ℝ^{m \times n}$ ,

𝑨 = [\begin{array}{cccc} 𝒇 (𝒆_{1}) & 𝒇 (𝒆_{2}) & \dots & 𝒇 (𝒆_{n}) \end{array}] = [\begin{array}{cccc} 𝒂_{1} & 𝒂_{2} & \dots & 𝒂_{n} \end{array}] .

$•$ $\circ$ Example 1. Rotation by $θ$ in $ℝ^{2}$ is obtained from

𝒇 (𝒆_{1}) = [\begin{array}{l} \cos θ \\ \sin θ \end{array}], f (𝒆_{2}) = [\begin{array}{l} - \sin θ \\ \cos θ \end{array}]

leading to

𝑨 = [\begin{array}{ll} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] .

∴	n=15; θ=2*π/n; c=cos(θ); s=sin(θ); A=[c -s; s c];

∴	v=[1; 0]; x=[0]; y=[0];

∴	for k=0:n global x,y,v,A x=[x v[1] 0]; y=[y v[2] 0] v = A*v end

∴	clf(); plot(x',y'); axis("equal");

∴

The column space of $𝑨$ is a vector subspace of the codomain, $C (𝑨) \leq ℝ^{m}$ , but according to the definition of dimension if $n < m$ there remain non-zero vectors within the codomain that are outside the range of $𝑨$ ,

n < m \Rightarrow \exists 𝒗 \in ℝ^{m}, 𝒗 \neq 𝟎, 𝒗 \notin C (𝑨) .

All of the non-zero vectors in $N (𝑨^{T})$ , namely the set of vectors orthogonal to all columns in $𝑨$ fall into this category. The above considerations can be stated as

\begin{array}{cccc} C (𝑨) \leq ℝ^{m}, & N (𝑨^{T}) \leq ℝ^{m}, & C (𝑨) ⊥ N (𝑨^{T}) & C (𝑨) + N (𝑨^{T}) \leq ℝ^{m} \end{array} .

The question that arises is whether there remain any non-zero vectors in the codomain that are not part of $C (𝑨)$ or $N (𝑨^{T})$ . The fundamental theorem of linear algebra states that there no such vectors, that $C (𝑨)$ is the orthogonal complement of $N (𝑨^{T})$ , and their direct sum covers the entire codomain $C (𝑨) \oplus N (𝑨^{T}) = ℝ^{m}$ .

Lemma 2. Let $𝒰, 𝒱$ , be subspaces of vector space $𝒲$ . Then $𝒲 = 𝒰 \oplus 𝒱$ if and only if

$𝒲 = 𝒰 + 𝒱$ , and

$𝒰 \cap 𝒱 = {𝟎}$ .

Proof. $𝒲 = 𝒰 \oplus 𝒱 \Rightarrow 𝒲 = 𝒰 + 𝒱$ by definition of direct sum, sum of vector subspaces. To prove that $𝒲 = 𝒰 \oplus 𝒱 \Rightarrow 𝒰 \cap 𝒱 = {𝟎}$ , consider $𝒘 \in 𝒰 \cap 𝒱$ . Since $𝒘 \in 𝒰$ and $𝒘 \in 𝒱$ write

$\begin{array}{llll} 𝒘 = 𝒘 + 𝟎 & (𝒘 \in 𝒰, 𝟎 \in 𝒱), & 𝒘 = 𝟎 + 𝒘 & (𝟎 \in 𝒰, 𝒘 \in 𝒱), \end{array}$

and since expression $𝒘 = 𝒖 + 𝒗$ is unique, it results that $𝒘 = 𝟎$ . Now assume (i),(ii) and establish an unique decomposition. Assume there might be two decompositions of $𝒘 \in 𝒲$ , $𝒘 = 𝒖_{1} + 𝒗_{1}$ , $𝒘 = 𝒖_{2} + 𝒗_{2}$ , with $𝒖_{1}, 𝒖_{2} \in 𝒰$ , $𝒗_{1}, 𝒗_{2} \in 𝒱 .$ Obtain $𝒖_{1} + 𝒗_{1} = 𝒖_{2} + 𝒗_{2}$ , or $𝒙 = 𝒖_{1} - 𝒖_{2} = 𝒗_{2} - 𝒗_{1}$ . Since $𝒙 \in 𝒰$ and $𝒙 \in 𝒱$ it results that $𝒙 = 𝟎$ , and $𝒖_{1} = 𝒖_{2}$ , $𝒗_{1} = 𝒗_{2}$ , i.e., the decomposition is unique. $□$

In the vector space $U + V$ the subspaces $U, V$ are said to be orthogonal complements is $U ⊥ V$ , and $U \cap V = {𝟎}$ . When $U \leq ℝ^{m}$ , the orthogonal complement of $U$ is denoted as $U^{⊥}$ , $U \oplus U^{⊥} = ℝ^{m}$ .

Theorem. Given the linear mapping associated with matrix $𝑨 \in ℝ^{m \times n}$ we have:

$C (𝑨) \oplus N (𝑨^{T}) = ℝ^{m}$ , the direct sum of the column space and left null space is the codomain of the mapping

$C (𝑨^{T}) \oplus N (𝑨) = ℝ^{n}$ , the direct sum of the row space and null space is the domain of the mapping

$C (𝑨) ⊥ N (𝑨^{T})$ and $C (𝑨) \cap N (𝑨^{T}) = {𝟎}$ , the column space is orthogonal to the left null space, and they are orthogonal complements of one another,
$\begin{array}{ll} C (𝑨) = N {(𝑨^{T})}^{⊥}, & N (𝑨^{T}) = C {(𝑨)}^{⊥} \end{array} .$

$C (𝑨^{T}) ⊥ N (𝑨)$ and $C (𝑨^{T}) \cap N (𝑨) = {𝟎}$ , the row space is orthogonal to the null space, and they are orthogonal complements of one another,
$\begin{array}{ll} C (𝑨^{T}) = N {(𝑨)}^{⊥}, & N (𝑨) = C {(𝑨^{T})}^{⊥} \end{array} .$

Consideration of equality between sets arises in proving the above theorem. A standard technique to show set equality

A = B

, is by double inclusion,

A \subseteq B \land B \subseteq A \Rightarrow A = B

. This is shown for the statements giving the decomposition of the codomain

ℝ^{m}

. A similar approach can be used to decomposition of

ℝ^{n}

The remainder of the FTLA is established by considering

𝑩 = 𝑨^{T}

, e.g., since it has been established in (v) that

C (𝑩) \oplus N (𝑨^{T}) = ℝ^{n}

, replacing

𝑩 = 𝑨^{T}

yields

C (𝑨^{T}) \oplus N (𝑨) = ℝ^{m}

, etc.

Fundamental Theorem of Linear Algebra

1.Partition of linear mapping domain and codomain