Chapter 6 extra note 0

Selected lecture notes

eigenvalue/eigenvector

Notation:

• $\mathcal{L}(V,W)$: The set of all linear transformation from vector space
  $V$ to the vector space
  $W$.
• $\mathcal{L}(V)$: The set of all linear transformation from vector space
  $V$ to the vector space
  $V$.
• $I$: the identity transformation.

Definition:
Let

$V$ be a vector space,

$T\in \mathcal{L}(V)$, a scalar

$\lambda \in F$ is called an eigenvalue if there exists a vector

$\mathbf{v}\in V\setminus \{\mathbf{0}\}$ such that

$T(\mathbf{v})=\lambda \mathbf{v}$. In addition,

$\mathbf{v}$ is called an eigenvector corresponding to

$\lambda$.

Proposition:
Let

$V$ be a (finite dimensional) vector space,

$T\in \mathcal{L}(V)$,

$\lambda \in F$, the followings are equivalent:

1. There exists
   $\mathbf{v}\in V\setminus \{\mathbf{0}\}$ such that
   $T(\mathbf{v})=\lambda \mathbf{v}$.
2. There exists
   $\mathbf{v}\in V\setminus \{\mathbf{0}\}$ such that
   $(T-\lambda I)(\mathbf{v})=\mathbf{0}$.
   • $T-\lambda I$ is not one-to-one.
   • $T-\lambda I$ is not onto.

     pf:
     Since
     

     $$\text{dim}(V)=\text{dim}(\text{Ker}(T-\lambda I))+\text{dim}(\text{Ran}(T-\lambda I)),$$
     and

     $\text{dim}(\text{Ker}(T-\lambda I))\ge 1$, so

     $\text{dim}(\text{Ran}(T-\lambda I))<n$ and

     $T-\lambda I$ is not onto.

3. $\text{Ker}(T-\lambda I)\ne \{\mathbf{0}\}$, or
   $\text{dim}(\text{Ker}(T-\lambda I))\ge 1$.
4. $T-\lambda I$ is not invertible.

Definition:
Let

$V$ be a vector space,

$T\in \mathcal{L}(V)$, the

$\lambda$-eigenspace is

$$V_{\lambda }=\{\mathbf{v}\in V|T(\mathbf{v})=\lambda \mathbf{v}\}.$$

Remark:

$\mathbf{0}\in V_{\lambda }$ for any

$\lambda \in F$.

Proposition:

• $V_{\lambda }$ is a vector subspace.
• $\text{Ker}(T)=V_0=\{\mathbf{v}\in V|T(\mathbf{v})=\mathbf{0}\}$.

Definition:
The geometric multiplicity of an eigenvalue

$\lambda$ of

$T$ is defined as

$\text{dim}(V_{\lambda })$.

Q: Does every linear transformation has an eigenvalue?

Example 1: (No eigenvalue)

Let

$V=\{\mathbf{0}\}$,

$T\in \mathcal{L}(V)$,

$T(\mathbf{v})=\mathbf{v}$.

Example 2: (No eigenvalue)

Let

$V={\mathbb{R}}^2$ with scalar field

$\mathbb{R}$. We define

$$T\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)=\left[\begin{array}{c}-y\\ x\end{array}\right].$$

Example 3: (Has eigenvalue)

Let

$V={\mathbb{C}}^2$ with scalar field

$\mathbb{C}$. We define

$$T\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)=\left[\begin{array}{c}-y\\ x\end{array}\right].$$

The eigenvalues are

$i$ and

$-i$.

Example 4: (Has eigenvalue)

Let

$V=C^1(\mathbb{R})$ (continuously differentiable functions) and define the linear transformation

$T(f)=f^{\prime }$, the first derivative of

$f$.

The eigenvalue

$\lambda$ should satisfy

$T(f)=f^{\prime }=\lambda f$, so we find the eigenvector

$f(x)=e^{\lambda x}$.

Therefore, we have

$\lambda \in (-\mathrm{\infty },\mathrm{\infty })$.

---

Theorem:
Let

$V$ be a vector space,

$T\in \mathcal{L}(V)$. Assume

$\{{\lambda }_1,\cdots ,{\lambda }_n\}$ are distinct eigenvalues with corresponding eigenvectors

$\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_n\}$. Then

$\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_n\}$ is linearly independent.

• Proof:

  We prove the statement by induction.

  Let

  $n=1$.

  Since

  ${\mathbf{v}}_1$ is an eigenvector, so

  ${\mathbf{v}}_1\ne \mathbf{0}$ and hence

  $\{{\mathbf{v}}_1\}$ is linearly independent.

  Assuming that

  $n=k$ is true, given

  $k$ distinct eigenvalues with

  $k$ corresponding eigenvectors, the eigenvectors form a linearly independent set.

  Consider

  $n=k+1$ and given

  $\{{\lambda }_1,\cdots ,{\lambda }_{k+1}\}$ distinct eigenvalues with corresponding eigenvectors

  $\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_{k+1}\}$.

  Since

  $\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_k\}$ is linearly independent, we only need to check whether or not

  ${\mathbf{v}}_{k+1}\in \text{span}\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_k\}$.

  Suppose
  

  $$\begin{array}{}\text{(1)}& {\mathbf{v}}_{k+1}={\alpha }_1{\mathbf{v}}_1+\cdots +{\alpha }_k{\mathbf{v}}_k.\end{array}$$
  Apply the transformation

  $T$ on (1) gives
  

  $$\begin{array}{}\text{(2)}& {\lambda }_{k+1}{\mathbf{v}}_{k+1}={\alpha }_1{\lambda }_1{\mathbf{v}}_1+\cdots +{\alpha }_k{\lambda }_k{\mathbf{v}}_k.\end{array}$$
  Multiplying (1) by

  ${\lambda }_{k+1}$ gives
  

  $$\begin{array}{}\text{(3)}& {\lambda }_{k+1}{\mathbf{v}}_{k+1}={\alpha }_1{\lambda }_{k+1}{\mathbf{v}}_1+\cdots +{\alpha }_k{\lambda }_{k+1}{\mathbf{v}}_k.\end{array}$$
  (2)-(3) then gives
  

  $$\mathbf{0}={\alpha }_1({\lambda }_1-{\lambda }_{k+1}){\mathbf{v}}_1+\cdots +{\alpha }_k({\lambda }_k-{\lambda }_{k+1}){\mathbf{v}}_k.$$
  Since the eigenvalues are distinct,

  ${\lambda }_i-{\lambda }_{k+1}\ne 0$ for all

  $i$. Also

  $\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_k\}$ is linearly independent, so we must have
  

  $${\alpha }_1=\cdots ={\alpha }_k=0.$$
  That is, use (1),

  ${\mathbf{v}}_{k+1}=\mathbf{0}$, which gives a contradiction as

  ${\mathbf{v}}_{k+1}$ is an eigenvector, should not be a zero vector.
  So

  ${\mathbf{v}}_{k+1}\notin \text{span}\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_k\}$ and

  $\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_{k+1}\}$ is linearly independent.

  Therefore, the statement is true by mathematical induction.

Proposition:
Suppose

$\text{dim}(V)=n$, then

$T\in \mathcal{L}(V)$ has at most

$n$ eigenvalues.

---

Definition:
Let

$V$ be a vector space,

$T\in \mathcal{L}(V)$ and

$m\in \mathbb{N}$. We define

$$T^m=\underset{m\text{ terms}}{\underbrace{T\circ T\circ \cdots \circ T}}.$$

Definition:
Let

$V$ be a vector space,

$T\in \mathcal{L}(V)$ and

$p(z)=a_0+a_{1}z+\cdots a_m{z}^m$ is a polynomial, we define a transformation

$p(T):V\to V$ as

$$p(T)=a_{0}I+a_{1}T+\cdots a_m{T}^m.$$

Proposition:

$p(T)\in \mathcal{L}(V)$.

Proposition:
Any two polynomial of an linear transformation commute, that is,

$$p(T)q(T)=q(T)p(T),$$
where

$p(z)$ and

$q(z)$ are polynomials.

Theorem:
Let

$V$ be a non-zero, finite dimensional complex vector space, let

$T\in \mathcal{L}(V)$, then

$T$ has at least one eigenvalue.

• Proof:

  Assume

  $\text{dim}(V)=n$, given

  $\mathbf{v}\in V\setminus \{\mathbf{0}\}$, then the set
  

  $$\{\mathbf{v},T\mathbf{v},\cdots ,T^n\mathbf{v}\}$$
  contains

  $n+1$ vectors and is a linearly dependent set. There exists

  $a_0,\cdots ,a_n$ not all zero such that
  

  $$\begin{array}{}\text{(4)}& a_0\mathbf{v}+a_{1}T\mathbf{v}+\cdots +a_n{T}^n\mathbf{v}=\mathbf{0}.\end{array}$$

  Let

  $m$ be such that

  $a_m\ne 0$ and

  $a_{m+1}=\cdots =a_n=0$.

  It is easy to see that

  $m\ge 1$.

  We define a polynomial

  $p(z)=a_0+a_{1}z+\cdots a_m{z}^m$. According to Fundamental theorem of Algebra, there exists

  ${\mu }_1,\cdots ,{\mu }_m\in \mathbb{C}$ such that
  

  $$\begin{array}{}\text{(5)}& p(z)=a_m(z-{\mu }_1)\cdots (z-{\mu }_m).\end{array}$$
  We can then rewrite (4) similarly as
  

  $$\begin{array}{rl}\mathbf{0}& =a_0\mathbf{v}+a_{1}T\mathbf{v}+\cdots +a_n{T}^n\mathbf{v}\\ & =p(T)\mathbf{v}\\ & =a_m(T-{\mu }_{1}I)\cdots (T-{\mu }_{m}I)\mathbf{v}.\end{array}$$
  Since

  $\mathbf{v}\ne \mathbf{0}$ and

  $a_m\ne 0$, there must be a

  $i\in \{1,\cdots ,m\}$ such that

  $\text{Ker}(T-{\mu }_{i}I)\ne \{\mathbf{0}\}$, that is,

  ${\mu }_i$ is an eigenvalue of

  $T$.

---

Let

$V$ be a non-zero, finite dimensional complex vector space, let

$T\in \mathcal{L}(V)$. Suppose

$\beta =\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_n\}\subset V$ is a basis, then we have a matrix representation of

$T$ as

$$A=[T{]}_{\beta }\in {\mathbb{M}}_{n\times n}.$$

Let

$\mathbf{x}=[x_1,\cdots ,x_n{]}^T\in {\mathbb{C}}^n$ be an eigenvector of

$A$ corresponding to the eigenvalue

$\lambda$, so that

$A\mathbf{x}=\lambda x$, we have

$\mathbf{v}=x_1{\mathbf{v}}_1+\cdots +x_n{\mathbf{v}}_n$ is an eigenvector of

$T$, and

$\lambda$ is an eigenvalue of

$T$.