LovisMind

第 22 章

对角化与矩阵的幂：把演化拆成互不干扰的方向

A = SΛS⁻¹ 让矩阵的幂变成特征值的幂；由此判断稳定性，并用差分方程与 Fibonacci 数列演示增长率的来源。

Diagonalization and Powers of A: Splitting Evolution into Independent Directions

A = SΛS⁻¹ turns powers of a matrix into powers of eigenvalues; this gives stability tests, solves difference equations, and explains the growth rate of Fibonacci numbers.

首发于 2009.05.07中 / EN
系统课程linear-algebradiagonalizationeigenvaluesdifference-equations
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对角化在做什么

上一讲得到 Ax=λxAx=\lambda x。这一讲把它组织成矩阵形式。把 nn 个线性无关的特征向量放进矩阵 SS 的列：

S=[x1x2⋯xn]S = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}

逐列计算 ASAS：

AS=[λ1x1λ2x2⋯λnxn]=SΛAS = \begin{bmatrix} \lambda_1 x_1 & \lambda_2 x_2 & \cdots & \lambda_n x_n \end{bmatrix} = S\Lambda

其中 Λ\Lambda 是以 λ1,…,λn\lambda_1,\dots,\lambda_n 为对角元的对角矩阵。左乘 S−1S^{-1}：

A=SΛS−1,Λ=S−1ASA = S\Lambda S^{-1}, \qquad \Lambda = S^{-1}AS

这就是对角化。它是与 A=LUA=LU、A=QRA=QR 并列的第三种分解，也是整个特征值章节的核心。

前提很清楚：SS 必须可逆，也就是需要 nn 个线性无关的特征向量。

时间轴

时间内容
0:10第二讲：对角化与矩阵的幂
3:16把 AS=SΛAS=S\Lambda 写成 A=SΛS−1A=S\Lambda S^{-1}
6:23对角化的条件：nn 个无关特征向量
9:28A2A^2 的特征值与特征向量
12:44Ak=SΛkS−1A^k=S\Lambda^kS^{-1}
15:50稳定性：所有 ∣λ∣<1\lvert\lambda\rvert<1
19:08特征值互异时一定可对角化
22:14不可对角化的情形
25:36重根导致的失败
28:40差分方程 uk+1=Auku_{k+1}=Au_k
31:41uk=SΛkS−1u0u_k=S\Lambda^kS^{-1}u_0
34:54Fibonacci 递推
37:59把递推化成二维系统
41:02特征值满足 λ2=λ+1\lambda^2=\lambda+1
44:16λ1≈1.618\lambda_1\approx1.618 主导增长
50:25一阶演化系统的一般配方

为什么对角化有用：幂

从定义直接看

若 Ax=λxAx=\lambda x，两边再乘 AA：

A2x=A(λx)=λAx=λ2xA^2x = A(\lambda x) = \lambda Ax = \lambda^2 x

所以 A2A^2 的特征值平方，特征向量不变。一般地：

Akx=λkxA^k x = \lambda^k x

从分解看

A2=(SΛS−1)(SΛS−1)=SΛ2S−1A^2 = (S\Lambda S^{-1})(S\Lambda S^{-1}) = S\Lambda^2 S^{-1}

中间的 S−1SS^{-1}S 消掉。对 kk 次幂：

Ak=SΛkS−1A^k = S\Lambda^k S^{-1}

而 Λk\Lambda^k 只是对角元各自取 kk 次方。

这正是对角化的价值。若用 LULU 计算 A100A^{100}，会得到一百个 LULU 相乘，无从下手；用特征分解，只是一个对角矩阵取幂。

稳定性定理

Ak→0A^k \to 0 当且仅当所有特征值的绝对值小于 1：

∣λi∣<1对所有 i|\lambda_i| < 1 \quad \text{对所有 } i

原因在于 SS 与 S−1S^{-1} 固定不动，只有 Λk\Lambda^k 在变化。绝对值小于 1 的数越乘越小；等于 1 不衰减；大于 1 会发散。特征值可以是负数或复数，所以必须取绝对值。

哪些矩阵可以对角化

情形是否可对角化说明
特征值互不相同一定可以随机矩阵几乎总属于这一类
特征值重复，但特征空间足够大可以例如单位矩阵，每个向量都是特征向量
特征值重复且特征向量不足不可以例如 [2102]\begin{bmatrix}2&1\\0&2\end{bmatrix}

重复特征值的两种命运

单位矩阵：特征值全是 1，重复 nn 次，但每个向量都是特征向量，可以选出 nn 个线性无关的。它本来就是对角的。

反例：

A=[2102]A=\begin{bmatrix}2&1\\0&2\end{bmatrix}

特征多项式是 (2−λ)2(2-\lambda)^2，所以 λ=2\lambda=2 重复两次。但

A−2I=[0100]A-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}

零空间只有一维，基是 (1,0)(1,0)。找不到第二个线性无关的特征向量，因此不可对角化。

这里引入两个概念：

  • 代数重数：特征值作为特征多项式根的重数，这里是 2。
  • 几何重数：对应特征空间的维数，这里是 1。

可对角化要求每个特征值的几何重数等于代数重数。

差分方程：每一步乘以 A

考虑一阶系统

uk+1=Auku_{k+1} = A u_k

给定初值 u0u_0，解是

uk=Aku0u_k = A^k u_0

这是最紧凑的写法，但并不能直接算出数值。真正的解法是把 u0u_0 展开成特征向量的组合：

u0=c1x1+c2x2+⋯+cnxn=Scu_0 = c_1 x_1 + c_2 x_2 + \cdots + c_n x_n = Sc

每乘一次 AA，每个分量独立地乘上自己的特征值：

uk=c1λ1kx1+c2λ2kx2+⋯+cnλnkxnu_k = c_1\lambda_1^k x_1 + c_2\lambda_2^k x_2 + \cdots + c_n\lambda_n^k x_n

写成矩阵形式就是 uk=SΛkcu_k = S\Lambda^k c，与 Aku0=SΛkS−1u0A^k u_0 = S\Lambda^k S^{-1}u_0 完全一致。

一句话总结： 特征向量把耦合的系统拆成 nn 个互不干扰的一维问题。

Fibonacci 数列：增长率的来源

把二阶递推变成一阶系统

Fk+2=Fk+1+FkF_{k+2} = F_{k+1} + F_k

这是二阶标量递推。引入向量

uk=[Fk+1Fk]u_k = \begin{bmatrix}F_{k+1}\\F_k\end{bmatrix}

并补一个恒等式 Fk+1=Fk+1F_{k+1}=F_{k+1}，得到

uk+1=[1110]uku_{k+1} = \begin{bmatrix}1&1\\1&0\end{bmatrix} u_k

记这个矩阵为 AA。

特征值

特征多项式：

det⁡(A−λI)=∣1−λ11−λ∣=λ2−λ−1\det(A-\lambda I) = \begin{vmatrix}1-\lambda&1\\1&-\lambda\end{vmatrix} = \lambda^2-\lambda-1

用求根公式：

λ=1±52\lambda = \frac{1\pm\sqrt5}{2}

所以

λ1=1+52≈1.618,λ2=1−52≈−0.618\lambda_1 = \frac{1+\sqrt5}{2}\approx 1.618, \qquad \lambda_2 = \frac{1-\sqrt5}{2}\approx -0.618

验证：和为 1（迹），积为 −1-1（行列式）。

注意 λ2−λ−1=0\lambda^2-\lambda-1=0 正是原递推的特征方程，这说明两种做法本质相同。

增长由谁决定

初始向量

u0=[F1F0]=[10]u_0=\begin{bmatrix}F_1\\F_0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}

把它写成两个特征向量的组合。此后

uk=c1λ1kx1+c2λ2kx2u_k = c_1\lambda_1^k x_1 + c_2\lambda_2^k x_2

由于 ∣λ1∣>1|\lambda_1|>1、∣λ2∣<1|\lambda_2|<1，第二项随 kk 增大迅速消失，第一项主导：

Fk≈c1(1+52)kF_k \approx c_1\left(\frac{1+\sqrt5}{2}\right)^k

这就是黄金比例出现在 Fibonacci 数列中的原因。增长率不是猜出来的，而是最大特征值的模。

特征向量

对 2×22\times2 情形，可以直接看出 [λ1]\begin{bmatrix}\lambda\\1\end{bmatrix} 是特征向量：

[1110][λ1]=[λ+1λ]=λ[λ1]\begin{bmatrix}1&1\\1&0\end{bmatrix} \begin{bmatrix}\lambda\\1\end{bmatrix} = \begin{bmatrix}\lambda+1\\\lambda\end{bmatrix} = \lambda\begin{bmatrix}\lambda\\1\end{bmatrix}

最后一步用了 λ2=λ+1\lambda^2=\lambda+1。

核心回顾

  • A=SΛS−1A=S\Lambda S^{-1}，前提是 nn 个线性无关的特征向量。
  • Ak=SΛkS−1A^k=S\Lambda^k S^{-1}，特征值取 kk 次方，特征向量不变。
  • Ak→0A^k\to0 当且仅当所有 ∣λi∣<1|\lambda_i|<1。
  • 特征值互异一定可对角化；重复时看几何重数是否等于代数重数。
  • 差分方程的解：把初值展开到特征向量上，让每个分量按自己的特征值演化。
  • Fibonacci 的增长率等于最大特征值 (1+5)/2(1+\sqrt5)/2。

常见误区

  1. 以为特征值互异是可对角化的必要条件。它只是充分条件。
  2. 只比较代数重数，不看几何重数。重复特征值未必短缺。
  3. 在稳定性判据里忘记取绝对值。负特征值的幂会交替变号，但同样需要 ∣λ∣<1|\lambda|<1。
  4. 把 A=SΛS−1A=S\Lambda S^{-1} 写成 S−1ΛSS^{-1}\Lambda S。记住「AA 乘 SS」，即 AS=SΛAS=S\Lambda。
  5. 认为 Fibonacci 增长是线性的。它是指数增长，底数为黄金比例。

理解检验

1. 由 AS=SΛAS=S\Lambda 推导 A=SΛS−1A=S\Lambda S^{-1}，指出哪一步需要 SS 可逆。

参考答案：从 AS=SΛAS=S\Lambda 出发，在右边乘以 S−1S^{-1}：

(AS)S−1=(SΛ)S−1⟹A=SΛS−1.(AS)S^{-1}=(S\Lambda)S^{-1}\quad\Longrightarrow\quad A=S\Lambda S^{-1}.

这一步需要 S−1S^{-1} 存在，也就是 SS 的列（nn 个特征向量）线性无关。若特征向量不足，SS 奇异，只能写成 AS=SΛAS=S\Lambda，无法解出 AA 的对角化形式。注意顺序：是 SΛS−1S\Lambda S^{-1}，不是 S−1ΛSS^{-1}\Lambda S。

2. 若 AA 的特征值是 2,0.52,0.5，AkA^k 会趋于零吗？

参考答案：不会。稳定性判据要求所有特征值满足 ∣λi∣<1|\lambda_i|<1。这里 ∣2∣>1|2|>1，对应方向上的分量按 2k2^k 增长，所以 AkA^k 在上该方向发散。0.5k→00.5^k\to0 只说明另一个方向衰减，不能决定整体。只有当全部特征值的模都小于 1 时才有 Ak→0A^k\to0。

3. 构造一个特征值重复但可对角化的矩阵，再构造一个不可对角化的。

参考答案：可对角化：A=[2002]=2IA=\begin{bmatrix}2&0\\0&2\end{bmatrix}=2I。特征值 22 的代数重数是 2，但每个非零向量都是特征向量，几何重数也是 2，S=IS=I 即可。不可对角化：B=[2102]B=\begin{bmatrix}2&1\\0&2\end{bmatrix}。特征值仍是二重的 22，但

B−2I=[0100]B-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}

的零空间只有 span⁡{(1,0)}\operatorname{span}\{(1,0)\}，几何重数 1 << 代数重数 2，找不到两个线性无关的特征向量。

4. Fibonacci 矩阵的第二个特征值为什么最终不重要？

参考答案：λ1=1+52≈1.618\lambda_1=\frac{1+\sqrt5}{2}\approx1.618，λ2=1−52≈−0.618\lambda_2=\frac{1-\sqrt5}{2}\approx-0.618。因为 ∣λ2∣<1|\lambda_2|<1，λ2k→0\lambda_2^k\to0；虽然 λ2<0\lambda_2<0 使符号交替，但幅度指数衰减。把 uk=c1λ1kx1+c2λ2kx2u_k=c_1\lambda_1^k x_1+c_2\lambda_2^k x_2 除以 λ1k\lambda_1^k，第二项是 c2(λ2/λ1)kx2c_2(\lambda_2/\lambda_1)^k x_2，而 ∣λ2/λ1∣<1|\lambda_2/\lambda_1|<1，所以相对贡献趋于零，增长由 λ1k\lambda_1^k 主导。

5. 一个 3×33\times3 矩阵的特征值是 1,1,11,1,1，能否断言它可对角化？

参考答案：不能。代数重数是 3，但可对角化要求几何重数也等于 3，即 N(A−I)N(A-I) 的维数为 3（等价于 A=IA=I）。例如 A=IA=I 可对角化；而

A=[110010001]A=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}

的 A−IA-I 只有一列非零，零空间维数为 2，几何重数 2<32<3，不可对角化。必须逐个特征值比较两个重数才能下结论。

与前后课程的关系

这一讲处理离散时间（差分方程）。下一讲把同样的思路搬到连续时间：把 λk\lambda^k 换成 eλte^{\lambda t}，把 AkA^k 换成矩阵指数 eAte^{At}。稳定性判据也从单位圆内变成左半平面。

What diagonalization does

The previous lecture produced Ax=λxAx=\lambda x. This one assembles those equations into a matrix identity. Put nn independent eigenvectors in the columns of SS:

S=[x1x2⋯xn]S = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}

Multiply column by column:

AS=[λ1x1λ2x2⋯λnxn]=SΛAS = \begin{bmatrix} \lambda_1 x_1 & \lambda_2 x_2 & \cdots & \lambda_n x_n \end{bmatrix} = S\Lambda

where Λ\Lambda is diagonal with entries λ1,…,λn\lambda_1,\dots,\lambda_n. Left-multiply by S−1S^{-1}:

A=SΛS−1,Λ=S−1ASA = S\Lambda S^{-1}, \qquad \Lambda = S^{-1}AS

This is diagonalization, the third decomposition in the course alongside A=LUA=LU and A=QRA=QR, and the centerpiece of the eigenvalue chapter.

The condition is explicit: SS must be invertible, which means nn linearly independent eigenvectors.

Timeline

TimeTopic
0:10Second lecture: diagonalization and powers
3:16Writing AS=SΛAS=S\Lambda as A=SΛS−1A=S\Lambda S^{-1}
6:23The condition: nn independent eigenvectors
9:28Eigenvalues and eigenvectors of A2A^2
12:44Ak=SΛkS−1A^k=S\Lambda^kS^{-1}
15:50Stability: all ∣λ∣<1\lvert\lambda\rvert<1
19:08Distinct eigenvalues guarantee diagonalization
22:14When diagonalization fails
25:36Failure caused by repeated eigenvalues
28:40Difference equations uk+1=Auku_{k+1}=Au_k
31:41uk=SΛkS−1u0u_k=S\Lambda^kS^{-1}u_0
34:54The Fibonacci recurrence
37:59Turning the recurrence into a 2D system
41:02The eigenvalues satisfy λ2=λ+1\lambda^2=\lambda+1
44:16λ1≈1.618\lambda_1\approx1.618 dominates
50:25The general recipe for first-order systems

Why it matters: powers

From the definition

If Ax=λxAx=\lambda x, multiply by AA again:

A2x=A(λx)=λAx=λ2xA^2x = A(\lambda x) = \lambda Ax = \lambda^2 x

So A2A^2 has the squared eigenvalues and the same eigenvectors. In general,

Akx=λkxA^k x = \lambda^k x

From the decomposition

A2=(SΛS−1)(SΛS−1)=SΛ2S−1A^2 = (S\Lambda S^{-1})(S\Lambda S^{-1}) = S\Lambda^2 S^{-1}

The inner S−1SS^{-1}S cancels. For the kk-th power,

Ak=SΛkS−1A^k = S\Lambda^k S^{-1}

and Λk\Lambda^k is just the diagonal matrix of kk-th powers.

This is the payoff. Computing A100A^{100} from LULU would mean multiplying a hundred LULU factors; eigen-decomposition reduces it to taking powers of scalars.

The stability theorem

Ak→0A^k \to 0 exactly when every eigenvalue has absolute value below 1:

∣λi∣<1for all i|\lambda_i| < 1 \quad \text{for all } i

The reason is that SS and S−1S^{-1} are fixed while only Λk\Lambda^k changes. Numbers below 1 shrink, numbers equal to 1 persist, numbers above 1 blow up. Eigenvalues may be negative or complex, so the absolute value is essential.

Which matrices diagonalize

CaseDiagonalizable?Explanation
Distinct eigenvaluesAlwaysA random matrix almost surely falls here
Repeated eigenvalues with enough eigenvectorsYesThe identity: every vector is an eigenvector
Repeated eigenvalues with too few eigenvectorsNoExample: [2102]\begin{bmatrix}2&1\\0&2\end{bmatrix}

Two fates for a repeated eigenvalue

The identity has eigenvalue 1 repeated nn times, but every vector is an eigenvector, so nn independent ones can be chosen. It is already diagonal.

The counterexample:

A=[2102]A=\begin{bmatrix}2&1\\0&2\end{bmatrix}

The characteristic polynomial is (2−λ)2(2-\lambda)^2, so λ=2\lambda=2 has multiplicity two. But

A−2I=[0100]A-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}

has a one-dimensional null space spanned by (1,0)(1,0). There is no second independent eigenvector, so the matrix cannot be diagonalized.

Two notions appear:

  • Algebraic multiplicity: the multiplicity as a root of the characteristic polynomial, here 2.
  • Geometric multiplicity: the dimension of the eigenspace, here 1.

Diagonalizability requires that for every eigenvalue, geometric multiplicity equals algebraic multiplicity.

Difference equations: multiply by AA each step

Consider the first-order system

uk+1=Auku_{k+1} = A u_k

With initial vector u0u_0, the solution is

uk=Aku0u_k = A^k u_0

Compact, but not directly computable. The real method is to expand u0u_0 in eigenvectors:

u0=c1x1+c2x2+⋯+cnxn=Scu_0 = c_1 x_1 + c_2 x_2 + \cdots + c_n x_n = Sc

Each application of AA multiplies each component by its own eigenvalue:

uk=c1λ1kx1+c2λ2kx2+⋯+cnλnkxnu_k = c_1\lambda_1^k x_1 + c_2\lambda_2^k x_2 + \cdots + c_n\lambda_n^k x_n

In matrix form this is uk=SΛkcu_k = S\Lambda^k c, identical to Aku0=SΛkS−1u0A^k u_0 = S\Lambda^k S^{-1}u_0.

One-sentence summary: Eigenvectors split a coupled system into nn independent one-dimensional problems.

Fibonacci: where the growth rate comes from

Turning a second-order recurrence into a first-order system

Fk+2=Fk+1+FkF_{k+2} = F_{k+1} + F_k

Introduce the vector

uk=[Fk+1Fk]u_k = \begin{bmatrix}F_{k+1}\\F_k\end{bmatrix}

and add the trivial equation Fk+1=Fk+1F_{k+1}=F_{k+1}:

uk+1=[1110]uku_{k+1} = \begin{bmatrix}1&1\\1&0\end{bmatrix} u_k

Call this matrix AA.

Eigenvalues

The characteristic polynomial:

det⁡(A−λI)=∣1−λ11−λ∣=λ2−λ−1\det(A-\lambda I) = \begin{vmatrix}1-\lambda&1\\1&-\lambda\end{vmatrix} = \lambda^2-\lambda-1

By the quadratic formula,

λ=1±52\lambda = \frac{1\pm\sqrt5}{2}

so

λ1=1+52≈1.618,λ2=1−52≈−0.618\lambda_1 = \frac{1+\sqrt5}{2}\approx 1.618, \qquad \lambda_2 = \frac{1-\sqrt5}{2}\approx -0.618

Check: the sum is 1 (the trace) and the product is −1-1 (the determinant).

Notice that λ2−λ−1=0\lambda^2-\lambda-1=0 is exactly the original recurrence in disguise; the two approaches agree.

What controls growth

The initial vector is

u0=[F1F0]=[10]u_0=\begin{bmatrix}F_1\\F_0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}

Write it as a combination of the two eigenvectors. Then

uk=c1λ1kx1+c2λ2kx2u_k = c_1\lambda_1^k x_1 + c_2\lambda_2^k x_2

Since ∣λ1∣>1|\lambda_1|>1 and ∣λ2∣<1|\lambda_2|<1, the second term decays quickly and the first dominates:

Fk≈c1(1+52)kF_k \approx c_1\left(\frac{1+\sqrt5}{2}\right)^k

This is why the golden ratio appears. The growth rate is the modulus of the largest eigenvalue, not a guess.

Eigenvectors

In the 2×22\times2 case one can spot that [λ1]\begin{bmatrix}\lambda\\1\end{bmatrix} works:

[1110][λ1]=[λ+1λ]=λ[λ1]\begin{bmatrix}1&1\\1&0\end{bmatrix} \begin{bmatrix}\lambda\\1\end{bmatrix} = \begin{bmatrix}\lambda+1\\\lambda\end{bmatrix} = \lambda\begin{bmatrix}\lambda\\1\end{bmatrix}

The last step uses λ2=λ+1\lambda^2=\lambda+1.

Core review

  • A=SΛS−1A=S\Lambda S^{-1}, provided there are nn independent eigenvectors.
  • Ak=SΛkS−1A^k=S\Lambda^k S^{-1}: eigenvalues are powered, eigenvectors are unchanged.
  • Ak→0A^k\to0 exactly when every ∣λi∣<1|\lambda_i|<1.
  • Distinct eigenvalues guarantee diagonalizability; repeats require checking multiplicities.
  • Difference equations: expand the initial vector in eigenvectors and let each component evolve independently.
  • Fibonacci growth equals the largest eigenvalue, (1+5)/2(1+\sqrt5)/2.

Common misconceptions

  1. Thinking distinct eigenvalues are necessary. They are sufficient, not necessary.
  2. Checking only algebraic multiplicity. Repeats are fine if the eigenspace is large enough.
  3. Dropping the absolute value in the stability test. Negative eigenvalues alternate in sign, but still need ∣λ∣<1|\lambda|<1.
  4. Writing A=S−1ΛSA=S^{-1}\Lambda S. The mnemonic is "AA multiplies SS": AS=SΛAS=S\Lambda.
  5. Calling Fibonacci growth linear. It is exponential, with base the golden ratio.

Check your understanding

1. Derive A=SΛS−1A=S\Lambda S^{-1} from AS=SΛAS=S\Lambda and identify where invertibility of SS is used.

Reference answer: Start from AS=SΛAS=S\Lambda and multiply on the right by S−1S^{-1}:

(AS)S−1=(SΛ)S−1⟹A=SΛS−1.(AS)S^{-1}=(S\Lambda)S^{-1}\quad\Longrightarrow\quad A=S\Lambda S^{-1}.

This uses the existence of S−1S^{-1}, i.e. the nn eigenvectors (columns of SS) are linearly independent. With too few eigenvectors, SS is singular and one can only write AS=SΛAS=S\Lambda, not a diagonalization of AA. Mind the order: it is SΛS−1S\Lambda S^{-1}, not S−1ΛSS^{-1}\Lambda S.

2. If the eigenvalues are 22 and 0.50.5, does AkA^k tend to zero?

Reference answer: No. The stability criterion requires ∣λi∣<1|\lambda_i|<1 for every eigenvalue. Here ∣2∣>1|2|>1, so the component along that eigenvector grows like 2k2^k and AkA^k diverges in that direction. The decay of 0.5k0.5^k in the other direction cannot save it. Only when all eigenvalues have modulus below 1 does Ak→0A^k\to0.

3. Construct one matrix with a repeated eigenvalue that diagonalizes and one that does not.

Reference answer: Diagonalizable: A=[2002]=2IA=\begin{bmatrix}2&0\\0&2\end{bmatrix}=2I. The eigenvalue 22 has algebraic multiplicity 2, but every nonzero vector is an eigenvector, so the geometric multiplicity is also 2 and S=IS=I works. Not diagonalizable: B=[2102]B=\begin{bmatrix}2&1\\0&2\end{bmatrix}. The eigenvalue 22 is still double, but

B−2I=[0100]B-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}

has nullspace span⁡{(1,0)}\operatorname{span}\{(1,0)\} of dimension 1, so the geometric multiplicity 1 is below the algebraic multiplicity 2 and no two independent eigenvectors exist.

4. Why does the second Fibonacci eigenvalue ultimately stop mattering?

Reference answer: λ1=1+52≈1.618\lambda_1=\frac{1+\sqrt5}{2}\approx1.618 and λ2=1−52≈−0.618\lambda_2=\frac{1-\sqrt5}{2}\approx-0.618. Since ∣λ2∣<1|\lambda_2|<1, λ2k→0\lambda_2^k\to0; the negative sign makes the term alternate, but its magnitude decays exponentially. Dividing uk=c1λ1kx1+c2λ2kx2u_k=c_1\lambda_1^k x_1+c_2\lambda_2^k x_2 by λ1k\lambda_1^k leaves c2(λ2/λ1)kx2c_2(\lambda_2/\lambda_1)^k x_2, and ∣λ2/λ1∣<1|\lambda_2/\lambda_1|<1, so the relative contribution vanishes and growth is governed by λ1k\lambda_1^k.

5. A 3×33\times3 matrix has eigenvalues 1,1,11,1,1. Can you conclude it is diagonalizable?

Reference answer: No. The algebraic multiplicity is 3, but diagonalizability requires the geometric multiplicity to be 3 as well, i.e. dim⁡N(A−I)=3\dim N(A-I)=3 (equivalently A=IA=I). For instance A=IA=I diagonalizes; but

A=[110010001]A=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}

has A−IA-I of rank 1, so dim⁡N(A−I)=2<3\dim N(A-I)=2<3 and it is not diagonalizable. One must compare the two multiplicities for each eigenvalue before concluding.

How this connects

This lecture treats discrete time. The next moves the same machinery to continuous time: replace λk\lambda^k with eλte^{\lambda t} and AkA^k with the matrix exponential eAte^{At}. The stability region changes from the unit disk to the left half-plane.

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