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Direct-sum decompositions

Let V be a linear space over field 𝔽 (which is either ℂ set of complex numbers, or ℝ. the set of real numbers, or ℚ, rational numbers) and f : V ⇾ V be a linear mapping, so f is endomorphism of V.

A subspace U ⊆ V is called an invariant subspace of f if f(U) ⊆ U. Then, restricting the domain to U we get a linear mapping f|U : U ⇾ U defined by
f|U(x) = f(x) for x ∈ U.
   
Example 1:    ■
End of Example 1
Recall the following definition from section on direct sums.
A family { Xi }i∈I, where I is some set (finite or not) of indexes, of subspaces of linear spaqce V is called independent, when a finite sum \( \displaystyle \quad \sum \mathbf{x}_j \quad \) of elements xj ∈ Xj, can vanish only if xj = 0 for all j.
As we shall mainly be concerned with finite families of subspaces, we shall restrict the index set to be finite, or ℕ. This is only a notational simplification.

   
Example 2:

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End of Example 2
Let X = {X₁, X₂, … , Xk} be a linearly independent family of subspaces of V and let W = X₁ ⊕ X₂ ⊕ ⋯ ⊕ Xk be the direct sum of them. As we know from section of Part 3, any x ∈ W is uniquely written as x = x₁ + x₂ + ⋯ + xk with x₁ ∈ X₁, x₂ ∈ X₂, … , xk ∈ Xk. For family of linear endomorphismss fi : Xi ⇾ Xi (i = 1, 2, … , k), we define the linear mapping f₁ ⊕ f₂ ⊕ ⋯ ⊕ fk : W ⇾ W by
\[ \left( f_1 \oplus f_2 \oplus \cdots \oplus f_k \right) (\mathbf{x}) = f_1 (\mathbf{x}_1 ) + f_2 (\mathbf{x}_2 ) + \cdots + f_k (\mathbf{x}_k ) . \]
We say that this endomorphism is the direct sum of linear mappings. Suppose that each subspace Xi is finite dimensional with basis βi and the linear mapping fi is represented by a matrix Ai = ⟧fi⟦ based on βi for i = 1, 2, … , k. Then, β = β₁ ∪ β₂ ∪ ⋯ ∪ βk is a basis of W, and on this basis, the linear mapping f₁ ⊕ f₂ ⊕ ⋯ ⊕ fk is represented by the matrix
\begin{equation} \label{EqDirect.1} \mathbf{A} = \begin{bmatrix} \mathbf{A}_1 & \mathbf{0} & \cdots & \mathbf{0} \\ \mathbf{0} & \mathbf{A}_2 & \cdots & \mathbf{0} \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{0} & \mathbf{0} & \cdots & \mathbf{A}_k \end{bmatrix} . \end{equation}

We call this block diagonal matrix \eqref{EqDirect.1} the direct sum of matrices, which is written as A = A₁ ⊕ A₂ ⊕ ⋯ ⊕ Ak.
Let f : W ⇾ W be a linear endomorphism such that each Xi is an invariant subspace of f for i = 1, 2, … , k. In this case we say that X decomposes f. Then, we have \[ f = \left. f\right\vert_{X_1} \oplus \left. f\right\vert_{X_2} \oplus \cdots \oplus \left. f\right\vert_{X_k} , \] and we call this expression the direct sum decomposition of f.
Example 3:    ■
End of Example 3

For a linear mapping f : V ⇾ V, we define fn : V ⇾ V nductively by

\[ f^n \stackrel{\tiny def}{=} \begin{cases} I , & \quad \mbox{if} \quad n = 0 , \\ f \circ f^{n-1} & \quad \mbox{for} \quad n=1, 2, \ldots , \end{cases} \]
where I is the identity mapping on V. If V is finite dimensional with basis β and A = ⟦f⟧β is the representation matrix of f on V, then An is a representation matrix of fn on V.

We define the subspaces K(n) and R(n) of V by

\[ K^{(n)} = \mbox{kernel}\left( f^n \right) \qquad \mbox{and} \qquad R^{(n)} = \mbox{Image}\left( f^n \right) \]
We have the inclusion relations
\begin{align*} \left\{ \mathbf{0} \right\} &= K^{(0)} \subseteq K^{(1)} \subseteq K^{(2)} \subseteq \cdots , \\ V &= R^{(0)} \supseteq R^{(1)} \supseteq R^{(2)} \supseteq \cdots . \end{align*}
Example 4:    ■
End of Example 4
    Let us define
\begin{equation} \label{EqDirect.1} K \stackrel{\tiny def}{=} \bigcup_{n\ge 0} K^{(n)} \qquad \mbox{and} \qquad R \stackrel{\tiny def}{=} \bigcap_{n\ge 0} R^{(n)} . \end{equation}
These two sets, K and K\R are invariant subsets of f. Suppose that V is a finite-dimensional space. Then there exist k and m such that
\begin{align*} K^{(k)} &= K^{(k+1)} = K^{(k+2)} = \cdots = K , \\ R^{(m)} &= R^{(m+1)} = R^{(m+2)} = \cdots = R . \end{align*}
Theorem 1: Let f : V ⇾ V a linear endomorphist on finite dimensional vector space V, and let K and R are kernel and range subspaces constructed according to Eq.\eqref{EqDirect.1}. Then V = K ⊕ R.
Let k₀ = max{k, m}. Then the linear mapping fR : R ⇾ R is also bijective. In particular, we see that \( \displaystyle \quad K \cap R = \mbox{kernel}\left( \left( \left. f \right\vert_R \right)^{k_0} \right) = \{ 0 \} . \quad \) Hence {K, R} is linearly independent. For any u ∈ V, set \( \displaystyle \quad \mathbf{v} = f^{(k_0 )} (\mayjbf{u}) . \quad \) Sincev &isin: R,
Example 5:    ■
End of Example 5
Corollary 1: Every singular matrix can be decomposed into two block matrix: \[ \mathbf{A} \sim \begin{bmatrix} \mathbf{R} & \mathbf{0} \\ \mathbf{0} & \mathbf{K}\end{bmatrix} . \]
Example 6:    ■
End of Example 6

Block matrix decomposition

 


  1. Beezer, R.A., A First Course in Linear Algebra, 2017.