Section01 Introduction to Eigenvalues

Replaces the complicit Matrix $\boldsymbol{A}$ with a value ($\lambda$)

• Eigenvectors: Certain expceptional vectors $\boldsymbol{x}$ in the same direction as $\boldsymbol{Ax}$. (When $\boldsymbol{A} = \boldsymbol{I}$ any vectors are the Eigenvectors of $\boldsymbol{I}$) ($\color{#D0104C}{\boldsymbol{x} \ne \boldsymbol{0}}$)
• Eigenvalues: The value has the same effect to eigenvectors as the matrix $\boldsymbol{A}$. ($\boldsymbol{Ax} = \lambda \boldsymbol{x}$)

Examples

• Question 1： $\boldsymbol{A} = \left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]$, Calculating $\boldsymbol{A}^{100}$

  1. Calculating eigenvalues ($\lambda$)

     • How? $$\begin{array}{ll} \because & \boldsymbol{Ax} = \lambda \boldsymbol{x} \\ \therefore & (\boldsymbol{A} - \lambda \boldsymbol{I})\boldsymbol{x} = \boldsymbol{0} \\ \because & \boldsymbol{x} \ne \boldsymbol{0} \\ \therefore & \boldsymbol{A} - \lambda \boldsymbol{I} \text{ is a singular matrix.} \\ \therefore & (\det \boldsymbol{A} - \lambda \boldsymbol{I}) = 0\\ \end{array}$$
     • Result of this question $$\begin{array}{ll} \because & \boldsymbol{A} = \left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]\\ \therefore & \left\vert \begin{matrix} 0.8 - \lambda & 0.3 \\ 0.2 & 0.7 - \lambda \end{matrix} \right\vert = (0.8 - \lambda)(0.7 - \lambda) - 0.2 \times 0.3 = 0\\ \therefore & \lambda = 1 \text{ or } \lambda = \frac{1}{2} \\ \end{array}$$

  2. Calculating eigenvectors ($\boldsymbol{x}$)

     • The meaning of eigenvectors in the perspective of space
       • eigenvectors is contained in the nullspace of $\boldsymbol{A} - \lambda \boldsymbol{I}$
     • Solutions of this question (one basis of nullspace) $$\begin{array}{ll} \boldsymbol{x}_{1} = \left[ \begin{matrix} 1.5 \\ 1 \end{matrix} \right] \\ \boldsymbol{x}_{2} = \left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] \end{array}$$

  3. When $\boldsymbol{A}$ is squared, the eigenvectors stay the same, the eigenvalues are squred. ($\color{#D0104C}{\boldsymbol{A}^{n}\boldsymbol{x} = \lambda^{n} \boldsymbol{x}}$ ($\boldsymbol{x}$ is eigenvectors.))

  4. Separating columns of $\boldsymbol{A}$ into eigenvectors
     • Why? $$\begin{array}{ll} \because & \boldsymbol{AA} = \boldsymbol{A} \left[ \begin{matrix} \boldsymbol{a}_{1} & \boldsymbol{a}_{2} & \cdots & \boldsymbol{a}_{n} \end{matrix} \right] = \left[ \begin{matrix} \boldsymbol{Aa}_{1} & \boldsymbol{Aa}_{2} & \cdots & \boldsymbol{Aa}_{n} \end{matrix} \right] \\ \because & \text{When vector } \boldsymbol{x} \text{ is eigenvector.}, \boldsymbol{A}^{n}\boldsymbol{x} = \lambda^{n}\boldsymbol{x} \\ \therefore & \text{Separate column vectors } \boldsymbol{a}_{1}, \boldsymbol{a}_{2}\cdots \text{ into eigenvectors } \boldsymbol{x}_{1}, \boldsymbol{x}_{2}, \cdots \\ \therefore & \boldsymbol{a}_{1} = \sum_{i=1}^{n}\boldsymbol{b}_{i}\boldsymbol{x}_{i} \\ \therefore & \boldsymbol{Aa}_{1} = \sum_{i=1}^{n}\lambda_{i}\boldsymbol{b}_{i}\boldsymbol{x}_{i} \Rightarrow \boldsymbol{A}^{n}\boldsymbol{a}_{1} = \sum_{i=1}^{n}\lambda^{n}_{i}\boldsymbol{b}_{i}\boldsymbol{x}_{i} \\ \end{array}$$
     • Solutions of this question $$\begin{array}{ll} \because & \boldsymbol{A} = \left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]; \boldsymbol{x}_{1} = \left[ \begin{matrix} 1.5 \\ 1 \end{matrix} \right]; \boldsymbol{x}_{2} = \left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] \\ \because & \boldsymbol{AA} = \boldsymbol{A} \left[ \begin{matrix} \boldsymbol{a}_{1} & \boldsymbol{a}_{2} \end{matrix} \right] = \left[ \begin{matrix} \boldsymbol{Aa}_{1} & \boldsymbol{Aa}_{2} \end{matrix} \right]\\ \therefore & \boldsymbol{a}_{1} = 0.4 \boldsymbol{x}_{1} - 0.2 \boldsymbol{x}_{2} \\ \therefore & \boldsymbol{Aa}_{1} = \boldsymbol{A}(0.4 \boldsymbol{x}_{1} - 0.2 \boldsymbol{x}_{2}) = 0.4\lambda_{1}\boldsymbol{x}_{1} - 0.2\lambda_{2}\boldsymbol{x}_{2}\\ \therefore & \boldsymbol{A}^{n}\boldsymbol{a}_{1} = \boldsymbol{A}^{n}(0.4 \boldsymbol{x}_{1} - 0.2 \boldsymbol{x}_{2}) = 0.4\lambda_{1}^{n}\boldsymbol{x}_{1} - 0.2\lambda_{2}^{n}\boldsymbol{x}_{2}\\ \therefore & \boldsymbol{A}^{99}\boldsymbol{a}_{1} = 0.4 \times 1^{99} \cdot \left[ \begin{matrix} 1.5 \\ 1 \end{matrix} \right] - 0.2 \times (\frac{1}{2})^{99} \cdot \left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] \approx 0.4 \left[ \begin{matrix} 1.5 \\ 1 \end{matrix} \right] = \left[ \begin{matrix} 0.6 \\ 0.4 \end{matrix} \right]\\ & \text{In the same way, } \boldsymbol{A}^{99}\boldsymbol{a}_{2} = \left[ \begin{matrix} 0.6 \\ 0.4 \end{matrix} \right]\\ \therefore & \boldsymbol{A}^{100} = \left[ \begin{matrix} \boldsymbol{A}^{99}\boldsymbol{a}_{1} & \boldsymbol{A}^{99}\boldsymbol{a}_{2} \end{matrix} \right] = \left[ \begin{matrix} 0.6 & 0.6 \\ 0.4 & 0.4 \end{matrix} \right]\\ \end{array}$$

• Example 2: The projection matrix $\boldsymbol{P} = \left[ \begin{matrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{matrix} \right]$ has eigenvectors $\lambda = 1$ and $\lambda = 0$

  • Markov Matrix: Each column of $\boldsymbol{P}$ adds to 1, so $\lambda = 1$ is an eigenvalue.
  • Singular Matrix: $\boldsymbol{P}$ is singular, so $\lambda = 0$ is an eigenvalue.
  • Symmetric: $\boldsymbol{P}$ is symmetric, so its eigenvectors $(1,1), (1,-1)$ are perpendicular.
  • The properties of projection matrix
    • When $\lambda = 1$, $\boldsymbol{Px}_{1} = \boldsymbol{x}$, so eigenvector correspond with $\lambda = 1$ is the column space of $\boldsymbol{P}$
    • When $\lambda = 0$, $\boldsymbol{Px}_{2} = \boldsymbol{0}$, so eigenvector correspond with $\lambda = 0$ is the nullspace of $\boldsymbol{P}$.
    • Any vector could be separated into two eigenvectors. $\boldsymbol{v} = b_{1}\boldsymbol{x}_{1} + b_{2}\boldsymbol{x}_{2}$. The projection matrix only persist the part in column space and drop the part in nullspace. $$\boldsymbol{Pv} = b_{1}\lambda_{1}x_{1} + b_{2}\lambda_{2}\boldsymbol{x}_{2} = b_{1}\boldsymbol{x}_{1} + \boldsymbol{0}$$

• Example 3 $\boldsymbol{R} = 2 \boldsymbol{P} - \boldsymbol{I}$

  • When a matrix is shifted by $\boldsymbol{I}$, each $\lambda$ is shifted by $1$. No changes in eigenvectors.
  • $\lambda_{1} = 2 - 1 = 1; \lambda_{2} = 0 -1 = -1$

The Euqation for the Eigenvalue

$$(\det \boldsymbol{A} - \lambda \boldsymbol{I}) = 0$$

Calculating Steps

1. Compute the determinants of $\boldsymbol{A} - \lambda \boldsymbol{I}$. With $\lambda$ subtracted along the diagonal, this determinant start with $\lambda^{n}$ or $-\lambda^{n}$. It is a polynomial in $\lambda$ of degrees $n$.
2. Find the roots of this polynomial, by solving $\det (\boldsymbol{A} - \lambda \boldsymbol{I}) = 0$. The $n$ roots are the $n$ eigenvalues of $\boldsymbol{A}$. They make $\boldsymbol{A} - \lambda \boldsymbol{I}$ singular.
3. For each eigenvalue $\lambda$, solve $(\boldsymbol{A} - \lambda \boldsymbol{I})\boldsymbol{x} = 0$ to find an eigenvector $\boldsymbol{x}$

Determinant and Trace

• If you add a row of $\boldsymbol{A}$ to another rows, or exchange rows, the eigenvalues usually change!

• The product of the $n$ eigenvalues equals the determinant. $$\prod_{i=1}^{n}\lambda_{i} = {\det \boldsymbol{A}}$$

  • Proof $$\begin{array}{ll} \because & \det (\boldsymbol{A} - \lambda\boldsymbol{I}) = 0 \\ \therefore & f(\lambda) = (-1)^{n}(\lambda - \lambda_{1})(\lambda - \lambda_{2})\cdots(\lambda - \lambda_{n}) \\ & = (\lambda_{1} - \lambda)(\lambda_{2} - \lambda) \cdots (\lambda_{n} - \lambda) \\ & \text{Suppose } \lambda = 0 \\ \therefore & f(0) = \det(\boldsymbol{A} - \boldsymbol{0}) = \det \boldsymbol{A} = \prod_{i=1}^{n}\lambda_{i} \\ \end{array}$$

• The sum of the $n$ eigenvalues equals the summ of the $n$ diagonal entries $$\sum_{i=1}^{n}\lambda_{i} = \sum_{i=1}^{n}a_{ii} = {\rm Tr}(\boldsymbol{A})$$

  • Proof $$\begin{array}{ll} \because & \det(\boldsymbol{A} - \lambda \boldsymbol{I}) = f(\lambda) \\ & = (-1)^{n}(\lambda^{n} - \color{#D0104C}{\sum_{i=1}^{n}{a_{ii}} \lambda^{n-1}} + \cdots + (-1)^{n}\det \boldsymbol{A}) \\ & = (-1)^{n}(\lambda - \lambda_{1})(\lambda - \lambda_{2}) \cdots (\lambda - \lambda_{n}) \\ & = (-1)^{n}(\lambda^{n} - \color{#D0104C}{\sum_{i=1}^{n}\lambda_{i} \lambda^{n-1}} + \cdots) \\ \therefore & \sum_{i=1}^{n}\lambda_{i} = \sum_{i=1}^{n}a_{ii} \\ \end{array}$$
  • $\sum_{i=i}^{n}a_{11}$ could be denoted by ${\rm Tr}(\boldsymbol{A})$ (The trace of $\boldsymbol{A}$)

Imaginary Eigenvalues

• The eigenvalues might not be real number.
• The eigenvalues of $\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$ is $i$ and $-i$. The sum of eigenvalues is $0$; the product of eigenvalues is $1$.
• The relationship between linear algebra and $i$
  1. $\boldsymbol{S}^{\text{T}} = \boldsymbol{S}$ is correspond with real number.
  2. $\boldsymbol{S}^{\text{T}} = - \boldsymbol{S}$ is correspond with complex number.

Eigenvalues of $\boldsymbol{AB}$ and $\boldsymbol{A} + \boldsymbol{B}$

• $\boldsymbol{A}$ and $\boldsymbol{B}$ share the same $n$ independent eigenvectors if and only if $\boldsymbol{AB} = \boldsymbol{BA}$
• If vector $\boldsymbol{x}$ is the eigenvector of $\boldsymbol{A}$ and $\boldsymbol{B}$ at the same time, then: $$\boldsymbol{ABx} = \boldsymbol{A}\beta\boldsymbol{x} = \beta \boldsymbol{Ax} = \lambda\beta \boldsymbol{x}$$

Cayley–Hamilton theorem

$$\prod_{i=1}^{n}(\boldsymbol{A} - \lambda_{i}\boldsymbol{I}) = \boldsymbol{0}$$

• proof $$\begin{array}{ll} & \text{Define } p(x) = \vert x \boldsymbol{I} - \boldsymbol{A} \vert = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{0} \\ \because & (x \boldsymbol{I} - \boldsymbol{A})^{-1} = \frac{1}{\vert x \boldsymbol{I} - \boldsymbol{A} \vert} \mathrm{adj}(x \boldsymbol{I} - \boldsymbol{A})\\ \therefore & \vert x \boldsymbol{I} - \boldsymbol{A} \vert \boldsymbol{I} = (x \boldsymbol{I} - \boldsymbol{A})\mathrm{adj}(x\boldsymbol{I} - \boldsymbol{A}) \\ & \text{Define } \mathrm{adj}(x \boldsymbol{I} - \boldsymbol{A}) = \boldsymbol{D}_{n-1}x^{n-1} + \boldsymbol{D}_{n-2}x^{n-2} + \cdots + \boldsymbol{D}_{0} \\ \therefore & (x \boldsymbol{I} - \boldsymbol{A})\mathrm{adj}(x \boldsymbol{I} - \boldsymbol{A}) \\ &= \boldsymbol{D}_{n-1}x^{n} + (\boldsymbol{D}_{n-2} - \boldsymbol{AD}_{n-1})x^{n-1} + (\boldsymbol{D}_{n-3}-\boldsymbol{AD}_{n-2})x^{n-2} + \cdots - \boldsymbol{AD}_{0} \\ \because & (x \boldsymbol{I} - \boldsymbol{A})\mathrm{adj}(x \boldsymbol{I} - \boldsymbol{A}) = \vert x \boldsymbol{I} - \boldsymbol{A} \vert \boldsymbol{I} \\ \therefore & \boldsymbol{D}_{n-1} = a_{n}\boldsymbol{I} \\ & \boldsymbol{D}_{k} - \boldsymbol{AD}_{k+1} = a_{k+1}\boldsymbol{I} \Rightarrow \boldsymbol{D}_{k} = \boldsymbol{AD}_{k+1} + a_{k+1}\boldsymbol{I}\\ & -\boldsymbol{AD}_{0} = a_{0} \boldsymbol{I} \\ \therefore & \boldsymbol{AD}_{0} + a_{0}\boldsymbol{I} = \boldsymbol{0} \\ & \Rightarrow\boldsymbol{A}(\boldsymbol{AD}_{1} + a_{1}\boldsymbol{I}) + \boldsymbol{a}_{0}\boldsymbol{I} = \boldsymbol{0} \\ & \Rightarrow \boldsymbol{A}^{2}(\boldsymbol{AD_{2}} + a_{2}\boldsymbol{I}) + a_{1}\boldsymbol{A} + a_{0} \boldsymbol{I} = \boldsymbol{0} \\ & \Rightarrow \cdots \\ & \Rightarrow \boldsymbol{A}^{n-1}(\boldsymbol{A}\boldsymbol{D}_{n-1} + a_{n-1} \boldsymbol{I}) + a_{n-2}\boldsymbol{A}^{n-2} + \cdots + a_{0}\boldsymbol{I} = \boldsymbol{0} \\ & \Rightarrow \boldsymbol{A}^{n}\boldsymbol{D}_{n-1} + a_{n-1}\boldsymbol{A}^{n-1} + a_{n-2}\boldsymbol{A}^{n-2} + \cdots + a_{0}\boldsymbol{I} = \boldsymbol{0} \\ & \Rightarrow a_{n}\boldsymbol{A}^{n} + a_{n-1} \boldsymbol{A}^{n-1} + \cdots + a_{0} \boldsymbol{I} = \boldsymbol{0} \\ \therefore & p(\boldsymbol{A}) = \boldsymbol{0} \\ \because & p(x) = (x - \lambda_{1})(x - \lambda_{2})\cdots(x-\lambda_{n}) = \prod_{i=1}^{n}(x - \lambda_{i}) \\ \therefore & p(\boldsymbol{A}) = (\boldsymbol{A} - \lambda_{1}\boldsymbol{I})(\boldsymbol{A} - \lambda_{2}\boldsymbol{I})\cdots(\boldsymbol{A} - \lambda_{n}\boldsymbol{I}) = \prod_{i=1}^{n}(\boldsymbol{A} - \lambda_{i}\boldsymbol{I}) = \boldsymbol{0} \\ \end{array}$$
• Properties
  • $(\boldsymbol{A} - \lambda_{k}\boldsymbol{I})$ is in the nullspace of $\prod_{i\ne k}(\boldsymbol{A} - \lambda_{i}\boldsymbol{I})$ ($p(x) = \prod_{i \ne k}{(\boldsymbol{A} - \lambda_{i}\boldsymbol{I})}(\boldsymbol{A} - \lambda_{k}\boldsymbol{I}) = \boldsymbol{0}$)

Properties from Exercises

1. Suppose eigenvalues of $\boldsymbol{A}$ is $\lambda_{1}, \lambda_{2}, \cdots$, then the eigenvalues of $\boldsymbol{A}^{2}$ is $\lambda_{1}^{2}, \lambda_{2}^{2}, \cdots$, the eigenvectors is unchanged.

   • proof $$\begin{array}{ll} \because & \boldsymbol{Ax} = \lambda\boldsymbol{x} \\ \therefore & \boldsymbol{A}^{2}\boldsymbol{x} = \boldsymbol{A}\lambda \boldsymbol{x} \\ \therefore & \boldsymbol{A}^{2}\boldsymbol{x} = \lambda^{2} \boldsymbol{x} \\ \therefore & \text{The eigenvectors are unchanged, the eigenvalues are } \lambda^{2} \\ \end{array}$$

2. Every $\lambda$ is in the circle around one or more diagonal entries $a_{ii}$: $\vert a_{ii} - \lambda \vert \le \sum_{i\ne j} a_{ij}$

   • proof $$\begin{array}{ll} \because & \det (\boldsymbol{A} - \lambda \boldsymbol{I}) = 0 \\ \therefore & \text{The matrix } \boldsymbol{A} - \lambda \boldsymbol{I} \text{ couldn't be a } \textbf{diagonally dominant matrix.} \\ \therefore & \text{Exist one or more } a_{ii} \text{ make } \vert a_{ii} - \lambda \vert \le \sum_{i \ne j}{a_{ij}} \\ \end{array}$$
3. Eigenvalues and Eigenvectors of Inverse Matrix ($\lambda_{\boldsymbol{A}^{-1}} = \frac{1}{\lambda_{\boldsymbol{A}}}$)
   • proof $$\begin{array}{ll} \because & \boldsymbol{Ax} = \lambda \boldsymbol{x} \\ \therefore & \boldsymbol{A}^{-1}\boldsymbol{Ax} = \boldsymbol{A}^{-1}\lambda \boldsymbol{x} \\ & \boldsymbol{x} = \lambda\boldsymbol{A}^{-1}\boldsymbol{x} \\ & \boldsymbol{A}^{-1}\boldsymbol{x} = \frac{1}{\lambda}\boldsymbol{x} \\ \therefore & \text{The eigenvectors are unchanged, the eigenvalues are } \frac{1}{\lambda} \\ \end{array}$$
4. When there are $n$ repeated eigenvalues, the nullspace of $\boldsymbol{A} - \lambda \boldsymbol{I}$ has $n$ dimension.
5. $x^{3} = 1 \Rightarrow (x-1)(x^{2} + x + 1) = 0 \Rightarrow \left\{ \begin{array}{ll} x = 1 \\ x = \frac{1}{2}(-1 + \sqrt{3} \mathrm{i}) \\ x = \frac{1}{2}(-1 - \sqrt{3} \mathrm{i}) \end{array} \right.$
6. $\boldsymbol{A}^{\text{T}}$ and $\boldsymbol{A}$ have the same eigenvalues, but their eigenvectors are transpose to each other. $$\begin{array}{ll} \because & \boldsymbol{Ax} = \lambda \boldsymbol{x}\\ \therefore & (\boldsymbol{Ax})^{\text{T}} = \lambda \boldsymbol{x}^{\text{T}} \\ \end{array}$$
7. Eigenvalues of $\boldsymbol{B}^{\text{T}}\boldsymbol{B}$ are $\lambda^{2}$.
8. When $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same eigenvalues and eigenvectors, then $\boldsymbol{A} = \boldsymbol{B}$.
   • proof $$\begin{array}{ll} & \text{Suppose a vector } \boldsymbol{v} = \sum_{i=1}^{n}c_{i}\boldsymbol{x}_{i}, \boldsymbol{x}_{i} \text{ is the eigenvectors of } \boldsymbol{A} \text{ and } \boldsymbol{B}. \\ \because & \boldsymbol{Av} = \boldsymbol{A}\sum_{i=1}^{n}c_{i}\boldsymbol{x}_{i} = \sum_{i=1}^{n} c_{i} \boldsymbol{A}x_{i} = \sum_{i=1}^{n}c_{i}\lambda_{i}\boldsymbol{x}_{i}\\ \therefore & \boldsymbol{Av} = \sum_{i=1}^{n}c_{i}\boldsymbol{Bx}_{i} = \boldsymbol{B}\sum_{i=1}^{n}c_{i}\boldsymbol{x}_{i} = \boldsymbol{Bv} \\ \therefore & \boldsymbol{Av} = \boldsymbol{Bv} \\ & \boldsymbol{Avv}^{-1} = \boldsymbol{Bvv}^{-1} \\ & \boldsymbol{A} = \boldsymbol{B} \end{array}$$
9. Suppose the block $\boldsymbol{B}$ has eigenvalues: $\lambda_{b1}, \lambda_{b2}, \cdots, \lambda_{bn}$, block $\boldsymbol{C}$ has eigenvalues: $\lambda_{c1},\lambda_{c2},\cdots,\lambda_{cn}$, block $\boldsymbol{D}$ has eigenvalues: $\lambda_{d1}, \lambda_{d2}, \cdots, \lambda_{dn}$. $\boldsymbol{A} = \left[ \begin{matrix} \boldsymbol{B} & \boldsymbol{C} \\ \boldsymbol{0} & \boldsymbol{D} \end{matrix} \right]$, eigenvalues of $\boldsymbol{A}$ are $\lambda_{b1},\lambda_{b2}, \cdots, \lambda_{bn}, \lambda_{d1},\lambda_{d2},\cdots,\lambda_{dn}$
   • proof $$\begin{array}{ll} \because & \boldsymbol{A} = \left[ \begin{matrix} \boldsymbol{B} & \boldsymbol{C} \\ \boldsymbol{0} & \boldsymbol{D} \end{matrix} \right]\\ \therefore & \boldsymbol{A} - \lambda \boldsymbol{I} = \left[ \begin{matrix} \boldsymbol{B} - \lambda \boldsymbol{I} & \boldsymbol{C} \\ \boldsymbol{0} & \boldsymbol{D} - \lambda \boldsymbol{I} \end{matrix} \right]\\ \therefore & \det (\boldsymbol{A} - \lambda \boldsymbol{I}) = (\det (\boldsymbol{B} - \lambda \boldsymbol{I}))(\det (\boldsymbol{D} - \lambda \boldsymbol{I})) = 0 \\ \therefore & \text{Solutions to } \det (\boldsymbol{B} - \lambda \boldsymbol{I}) = 0 \text{ and solutions to } \det (\boldsymbol{D} - \lambda \boldsymbol{I}) = 0 \\ & \text{are solutions to } \det (\boldsymbol{A} - \lambda \boldsymbol{I}) = 0 \\ \therefore & \{ \lambda_{b1}, \lambda_{b2},\cdots, \lambda_{bn}, \lambda_{d1}, \lambda_{d2},\cdots, \lambda_{dn} \} \text{ are solutions to } \det (\boldsymbol{A} - \lambda \boldsymbol{I}) = 0 \\ \therefore & \text{Eigenvalues of } \boldsymbol{A} \text{ are the Union of eigenvalues of } \boldsymbol{B} \text{ and } \boldsymbol{D} \\ \end{array}$$
10. Suppose permutation matrix $\boldsymbol{P}$, the matrix $\boldsymbol{P}\boldsymbol{A}\boldsymbol{P}^{\text{T}}$ has the same eigenvalues with $\boldsymbol{A}$. (When do the same column exchange and row exchange, the eigenvalues are unchanged)
    • proof $$\begin{array}{ll} \because & \boldsymbol{Ax} = \lambda \boldsymbol{x} \\ \therefore & \boldsymbol{PAxP}^{\text{T}} = \boldsymbol{P}\lambda \boldsymbol{xP}^{\text{T}} \\ \because & \boldsymbol{P} \text{ is orthonormal matrix.} \\ \therefore & \boldsymbol{PP}^{\text{T}} = \boldsymbol{P}^{\text{T}}\boldsymbol{P} = \boldsymbol{I} \\ \therefore & \boldsymbol{PAxP}^{\text{T}} = (\boldsymbol{PAP}^{\text{T}}) (\boldsymbol{PxP}^{\text{T}}) = \lambda (\boldsymbol{PxP}^{\text{T}}) \\ \therefore & \text{When exchange same rows and columns at the same time. } \\ & \text{Eigenvalues of matrix are unchanged.} \end{array}$$
11. Suppose two vectors $\boldsymbol{u}$ and $\boldsymbol{v}$, $\boldsymbol{A} = \boldsymbol{u}\boldsymbol{v}^{\text{T}}$, $\boldsymbol{u}$ is a eigenvector of $\boldsymbol{A}$
12. proof $$\begin{array}{ll} \because & \boldsymbol{A} = \boldsymbol{u} \boldsymbol{v}^{\text{T}} \\ \therefore & \boldsymbol{Au} = \boldsymbol{u}\boldsymbol{v}^{\text{T}}\boldsymbol{u} = \boldsymbol{u}(\boldsymbol{v}^{\text{T}}\boldsymbol{u}) = \boldsymbol{u}(\boldsymbol{v}\cdot \boldsymbol{u}) = \boldsymbol{u}\lambda \\ \therefore & \boldsymbol{u} \text{ is an eigenvector of } \boldsymbol{uv}^{\text{T}}, \\ & \text{the corresponding eigenvalue is } \boldsymbol{v} \cdot \boldsymbol{u} \\ \end{array}$$