[Axler] The Existence of Eigenvalue

Eigenvectors and Upper-Triangular Matrices (Part 1)

Based on Linear Algebra Done Right by Sheldon Axler

This post covers the foundational concepts required to understand eigenvalues and eigenvectors, including operator polynomials, and presents the proof for the existence of eigenvalues over complex vector spaces.

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1. Notation and Terminology

• $\mathbf{F}$ : Denotes either the real field $\mathbf{R}$ or the complex field $\mathbf{C}$.
• $V$ : A vector space over $\mathbf{F}$.
• Operator : A linear map from a vector space to itself ($T: V \to V$).
• $\mathcal{L}(V)$ : The set of all operators on $V$.

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2. Powers of an Operator

For an operator $T \in \mathcal{L}(V)$ and a positive integer $m$:

• $T^m = \underbrace{T \circ \dots \circ T}_{m \text{ times}}$ (Composition of maps).
• $T^0 = I$ (Identity operator).
• If $T$ is invertible, $T^{-m} = (T^{-1})^m$.

Exponent Rules:
$T^m \circ T^n = T^{m+n}$
$(T^m)^n = T^{mn}$

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3. Polynomials Applied to Operators

Let $p(z) = a_0 + a_1 z + \dots + a_m z^m$ be a polynomial with coefficients in $\mathbf{F}$. We define $p(T)$ as:

$$ p(T) = a_0 I + a_1 T + \dots + a_m T^m $$

Example: Differentiation Operator
Let $P(\mathbf{R})$ be the space of real polynomials and let $D$ be the differentiation operator ($Dp = p'$).
If $p(x) = 7 - 3x + 5x^2$, then according to the definition:
$$ p(D) = 7I - 3D + 5D^2 $$ Applying this to a polynomial $q$: $$ p(D)(q) = 7q - 3q' + 5q'' $$

Algebraic Properties

The map $p \mapsto p(T)$ is a linear map from $P(\mathbf{F})$ to $\mathcal{L}(V)$. A crucial property is multiplicativity:

$$ (pq)(T) = p(T) \circ q(T) $$

Corollary (Commutativity):
Any two polynomials in $T$ commute with each other. $$ p(T) \circ q(T) = q(T) \circ p(T) $$ Since operator multiplication is generally not commutative, this property is very useful.

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4. Existence of Eigenvalues

Theorem

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

Important Constraints

• Real Vector Spaces: False. Consider rotation by $90^\circ$ on $\mathbf{R}^2$ ($T(x, y) = (-y, x)$). It has no real eigenvalues because no non-zero vector is mapped to a scalar multiple of itself.
• Infinite-dimensional Spaces: False. Consider the multiplication operator on complex polynomials defined by $(Tp)(z) = z \cdot p(z)$.

Proof (Without Determinants)

Let $V$ be a complex vector space with $\dim V = n > 0$, and $T \in \mathcal{L}(V)$.

1. Choose a nonzero vector $v \in V$.
2. Consider the list of $n+1$ vectors: $(v, Tv, T^2v, \dots, T^n v)$.
3. Since $\dim V = n$, this list is linearly dependent. Thus, there exist scalars $a_0, \dots, a_n \in \mathbf{C}$ (not all zero) such that: $$ a_0 v + a_1 Tv + \dots + a_n T^n v = 0 $$
4. Let $p(z) = a_0 + a_1 z + \dots + a_n z^n$. By the Fundamental Theorem of Algebra, we can factor $p(z)$: $$ p(z) = c(z - \lambda_1) \dots (z - \lambda_m) $$
5. Substituting $T$ for $z$, the equation becomes: $$ c(T - \lambda_1 I) \dots (T - \lambda_m I)v = 0 $$
6. Since $c \ne 0$ and $v \ne 0$, the operator product applied to $v$ is zero. This implies that at least one of the factors, say $(T - \lambda_j I)$, is not injective.
7. If $(T - \lambda_j I)$ is not injective, then its null space is non-trivial. Thus, $\lambda_j$ is an eigenvalue. $\blacksquare$

Note on Determinants:
Most textbooks prove this using the characteristic polynomial $\det(\lambda I - T)$. This book avoids that approach to define eigenvalues directly from the geometry and structure of vector spaces, without relying on the complex definition of determinants early on.