Table of Contents
Table of Contents
Invariant Subspaces
Definition 5.1 – Invariant Subspace
Let $T \in \mathcal{L}(V)$ be a (linear) operator on vector space $V$. When we restrict $T$ to a subspace $U \subseteq V$, we are not sure the corresponding vectors will still lie in $U$. If they do, we say that $U$ is an invariant subspace under $T$. We call $T|_U$ is an operator on $U$.
Considering one special invariant subspace, that is, one-dimensional invariant subspace, which is spanned by only one vector $v \in V$. If this one-dimensional subspace is invariant under $T$, then the linear map $T$ only depends on where the basis vector $v$ is mapped, which leads to the concept of eigenvalues and eigenvectors: $$ T v = \lambda v $$
Eigenvalues
Definition 5.2 – Eigenvalue and Eigenvector
Let $T \in \mathcal{L}(V)$ be a (linear) operator on vector space $V$. A scalar $\lambda \in \mathbb{F}$ is called an eigenvalue of $T$ if there exists a non-zero vector $v \in V$ such that $$ T v = \lambda v. $$ The vector $v$ is called an eigenvector corresponding to the eigenvalue $\lambda$.
Lemma 5.1 – Equivalence of Eigenvalue
Suppose $V$ is a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and $\lambda \in \mathbb{F}$. The following statements are equivalent:
-
$\lambda$ is an eigenvalue of $T$.
-
$T – \lambda I$ is not injective.
-
$T – \lambda I$ is not surjective.
-
$\det(T – \lambda I) = 0$.
Proof
Regarding $(1) \iff (2)$, by $T v = \lambda v$, we have $(T – \lambda I_V) v = 0$. Since by linearity of vector space of $\mathcal{L}(V)$, $T – \lambda I_V$ is also a operator on $V$, and its kernel is non-trivial as basis vector $v$ is non-zero. Completing the proof of $(1) \iff (2)$.
By the properties of invertible linear map of the same dimension $$ \text{injectivity} \iff \text{surjectivity} \iff \text{bijectivity} $$ , we have $(2) \iff (3) \iff (4)$, completing the proof.
Lemma 5.2 – Linearly Independent Eigenvectors
Eigenvectors corresponding to distinct eigenvalues of a linear operator are linearly independent.
Proof
Suppose $v_1, v_2, \dots, v_k$ are eigenvectors corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_k$ of a linear operator $T$. We prove the statement by induction on $k$.
For $k = 1$, the statement is trivial since a single non-zero vector is linearly independent.
Assume the statement holds for $k-1$ eigenvectors (i.e. if any linear combination of $k – 1$ distinct eigenvectors equals zero, then all coefficients are zeros). Consider $k$ eigenvectors $v_1, v_2, \dots, v_k$ corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_k$. Suppose $$ a_1 v_1 + a_2 v_2 + \dots + a_k v_k = 0. $$ Applying $T$ to both sides, we get $$ a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 + \dots + a_k \lambda_k v_k = 0. $$ Multiplying the first equation by $\lambda_k$ and subtracting from the second equation, we obtain $$ a_1 (\lambda_1 – \lambda_k) v_1 + a_2 (\lambda_2 – \lambda_k) v_2 + \dots + a_{k-1} (\lambda_{k-1} – \lambda_k) v_{k-1} = 0. $$ By the induction hypothesis, since $\lambda_i \ne \lambda_j$ for all $i \ne j$, we have $a_1 = a_2 = \dots = a_{k-1} = 0$. Substituting back into the original equation, we get $a_k v_k = 0$, which implies $a_k = 0$ since $v_k \neq 0$.
Hence, $v_1, v_2, \dots, v_k$ are linearly independent.
Lemma 5.3 – Maximum Number of Eigenvalues
A linear operator on an $n$-dimensional vector space has at most $n$ distinct eigenvalues.
Proof
By lemma 5.2, eigenvectors corresponding to distinct eigenvalues are linearly independent. Since a linear operator on an $n$-dimensional vector space cannot have more than $n$ linearly independent vectors, it follows that the number of distinct eigenvalues is at most $n$.
Polynomials Applies to Operators
Operators is more poppular than general linear maps, owe that the basis of domain and codomain are the same, thus invertibility is well-defined. Therefore, we can composite one operator many times, more in general we can apply any finite-degree polynomials to them directly.
Definition 5.3 – Polynomial Evaluation on Operator
Let $T$ be a linear operator on a vector space $V$, and let $p(x) = a_0 + a_1 x + \dots + a_m x^m$ be a polynomial. The polynomial evaluation of $p$ on $T$ is defined as $$ p(T) = a_0 I + a_1 T + \dots + a_m T^m, $$ where $I$ is the identity operator on $V$. For simplicity, we call $p(T)$ as polynomial operator on $T$.
Lemma 5.4 – Properties of Polynomial Operators
Due to the naive invertibility property of operators, the commutativity of operators with themselves ensures that polynomial evaluations on operators are well-defined and behave as expected. Suppose $p, q \in \mathcal{P}(\mathbb{F})$ are two polynomials, then we have $$ \begin{aligned} \text{(a) } &p(T) + q(T) = (p+q)(T), \\ \text{(b) } &p(T) q(T) = (pq)(T), \\ \text{(c) } &p(T) q(T) = q(T) p(T). \end{aligned} $$
Lemma 5.5 – Invariant of Null Space and Range of Polynomial Operators
Suppose $T$ is a linear operator on a vector space $V$, and $p(x)$ is a polynomial. Then the null space and range of $p(T)$ are invariant under $T$, i.e., $$ \begin{aligned} \text{(a) } &T(\text{null}(p(T))) \subseteq \text{null}(p(T)) \\ \text{(b) } &T(\text{range}(p(T))) \subseteq \text{range}(p(T)). \end{aligned} $$
Proof
Regarding (a), it is suffices to show that $\forall x \in T(\text{null}(p(T))) \to x \in \text{null}(p(T))$, which is equivalent with $\forall x \in \text{null}(p(T)) \to T(x) \in \text{null}(p(T))$.
By $p(T) \in \mathcal{L}(V)$, we have $p(T)(x) = 0$. And $T \in \mathcal{L}(V)$, we have $T(x) \in V$, by the commutativity of polynomial operators, we have: $$ p(T) \circ T(x) = (p(T) \circ T)(x) = (T \circ p(T))(x) = T(p(T)(x)) = T(0) = 0. $$ Thus $T(x) \in \text{null}(p(T))$.
Regarding (b), it is suffices to show that $\forall y \in \text{range}(p(T)) \to T(y) \in \text{range}(p(T))$.
Let $y \in \text{range}(p(T))$, then there exists $x \in V$ such that $y = p(T)(x)$. By the commutativity of polynomial operators, we have: $$ T(y) = T(p(T)(x)) = (T \circ p(T))(x) = (p(T) \circ T)(x) = p(T)(T(x)) \in \text{range}(p(T)). $$
