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第 24 章

Markov 矩阵与傅里叶级数：特征值的两个应用

列和为 1 的非负矩阵必有特征值 1，其正特征向量给出稳态；把同样的「正交基展开」推广到函数空间，就得到傅里叶系数。

Markov Matrices and Fourier Series: Two Uses of Eigenvalues

A nonnegative matrix whose columns sum to one always has eigenvalue 1, and its positive eigenvector is the steady state; the same orthonormal-basis expansion, moved to function space, gives Fourier coefficients.

首发于 2019.09.25中 / EN
系统课程linear-algebramarkov-matricesfourier-serieseigenvalues
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Markov 矩阵的定义

一个矩阵称为 Markov 矩阵，如果它满足：

  1. 所有元素非负；
  2. 每一列的元素之和为 1。

第二条在转置写法中会变成「每一行之和为 1」。本课程用列向量，因此约定列和为一。

课堂上随手写了一个例子：

A=[0.10.010.30.20.990.30.700.4]A=\begin{bmatrix} 0.1&0.01&0.3\\ 0.2&0.99&0.3\\ 0.7&0&0.4 \end{bmatrix}

每列相加都是 1。由于非负性与列和为 1 在矩阵乘法下保持，AA 的所有幂仍是 Markov 矩阵。

时间轴

时间内容
0:00马尔可夫矩阵与傅里叶级数
3:05回顾 λ=0\lambda=0 的稳态例子
6:45uku_k 的 λk\lambda^k 分解
9:56特征值 1 与衰减项
12:57马尔可夫矩阵：每列和为 1
16:01列相关，因此存在左零向量
19:05左特征向量与稳态
22:29人口流动的转移矩阵
25:40长期分布的计算
28:46第二个特征值 0.7
31:48uku_k 的完整分解
35:10转向投影与傅里叶
38:13用正交基展开向量
41:25把函数也当作向量
44:30傅里叶级数的系数
47:36系数计算的例子

关键结论

结论内容
1λ=1\lambda=1 一定是特征值
2其余特征值满足 $
3λ=1\lambda=1 对应的特征向量所有分量非负，给出稳态

为什么 1 一定是特征值

考虑 A−IA-I。由于 AA 每列之和为 1，A−IA-I 的每列之和为 0。换句话说，把 A−IA-I 的三行相加，得到零行，因此行线性相关，矩阵奇异。

A−I 奇异  ⟺  det⁡(A−I)=0  ⟺  λ=1 是特征值A-I \text{ 奇异} \iff \det(A-I)=0 \iff \lambda=1 \text{ 是特征值}

还有另一种说法：全 1 向量 1\mathbf{1} 满足

1T(A−I)=0\mathbf{1}^T(A-I)=0

所以 1\mathbf{1} 在 A−IA-I 的左零空间里。这个向量不是 AA 的特征向量，而是 ATA^T 的特征向量。由于 AA 与 ATA^T 的特征值相同（det⁡(AT−λI)=det⁡(A−λI)\det(A^T-\lambda I)=\det(A-\lambda I)），这同样证明了 1 是 AA 的特征值。

稳态从哪里来

差分方程 uk+1=Auku_{k+1}=Au_k 的解为

uk=c11kx1+c2λ2kx2+⋯u_k=c_1 1^k x_1+c_2\lambda_2^k x_2+\cdots

当 ∣λi∣<1|\lambda_i|<1 时，除第一项外全部衰减到零，于是

uk→c1x1u_k\to c_1x_1

稳态就是 λ=1\lambda=1 的特征向量（乘一个由初始总量决定的系数）。

人口流动的例子

设只有两个州：加州（C）与马萨诸塞州（M）。每年：

  • 加州 90% 留下，10% 迁往麻省；
  • 麻省 80% 留下，20% 迁往加州。
[uCuM]k+1=[0.90.20.10.8][uCuM]k\begin{bmatrix}u_C\\u_M\end{bmatrix}_{k+1} = \begin{bmatrix}0.9&0.2\\0.1&0.8\end{bmatrix} \begin{bmatrix}u_C\\u_M\end{bmatrix}_{k}

矩阵每列和为 1，总人口守恒。

特征值

迹为 1.71.7，所以另一个特征值是 0.70.7：

λ1=1,λ2=0.7\lambda_1=1,\qquad \lambda_2=0.7

特征向量

λ=1\lambda=1：解 (A−I)x=0(A-I)x=0，其中

A−I=[−0.10.20.1−0.2]A-I=\begin{bmatrix}-0.1&0.2\\0.1&-0.2\end{bmatrix}

零空间由 (2,1)(2,1) 张成。

λ=0.7\lambda=0.7：解 (A−0.7I)x=0(A-0.7I)x=0，得 (1,−1)(1,-1)。

具体演化

设初始 u0=(0,1000)u_0=(0,1000)，即所有人都在麻省。写成特征向量的组合：

[01000]=c1[21]+c2[1−1]\begin{bmatrix}0\\1000\end{bmatrix} = c_1\begin{bmatrix}2\\1\end{bmatrix} +c_2\begin{bmatrix}1\\-1\end{bmatrix}

解得

c1=10003,c2=−20003c_1=\frac{1000}{3},\qquad c_2=-\frac{2000}{3}

于是

uk=10003[21]−20003(0.7)k[1−1]u_k=\frac{1000}{3}\begin{bmatrix}2\\1\end{bmatrix} -\frac{2000}{3}(0.7)^k\begin{bmatrix}1\\-1\end{bmatrix}

当 k→∞k\to\infty，第二项消失：

u∞=[2000/31000/3]u_\infty=\begin{bmatrix}2000/3\\1000/3\end{bmatrix}

总人数仍是 1000。第一步验算：u1=Au0=(200,800)u_1=A u_0=(200,800)，与公式一致。

补充： 这个例子假设转移概率每年不变，且人口只在两个州之间流动。真实人口模型还会有出生、死亡与跨州迁移，那时矩阵不再守恒。

投影到正交基：傅里叶级数的伏笔

设 q1,…,qnq_1,\dots,q_n 是 nn 维空间的一组标准正交基。任意向量 vv 可以展开为

v=x1q1+x2q2+⋯+xnqnv=x_1q_1+x_2q_2+\cdots+x_nq_n

两边与 qiq_i 做内积。由于正交性，只有第 ii 项存活：

qiTv=xi(qiTqi)=xiq_i^Tv=x_i(q_i^Tq_i)=x_i

所以

xi=qiTvx_i=q_i^Tv

矩阵形式是 Qx=vQx=v，而 Q−1=QTQ^{-1}=Q^T，于是 x=QTvx=Q^Tv。这正是「正交基让求系数变成一次内积」的意思。

从向量到函数：傅里叶级数

傅里叶的洞见是把上面的结构搬到函数空间：

  • 向量 vv 换成函数 f(x)f(x)；
  • 基向量 qiq_i 换成函数 1,cos⁡x,sin⁡x,cos⁡2x,sin⁡2x,…1,\cos x,\sin x,\cos 2x,\sin 2x,\dots；
  • 有限维换成无限维。

函数的内积

向量内积是「逐分量相乘再相加」。函数有无穷多个点，求和变成积分：

⟨f,g⟩=∫02πf(x)g(x) dx\langle f,g\rangle=\int_0^{2\pi} f(x)g(x)\,dx

积分区间取 [0,2π][0,2\pi]，因为所有基函数都以 2π2\pi 为周期。

正交性

以 sin⁡x\sin x 与 cos⁡x\cos x 为例：

∫02πsin⁡xcos⁡x dx=12∫02πsin⁡2x dx=0\int_0^{2\pi}\sin x\cos x\,dx = \frac12\int_0^{2\pi}\sin 2x\,dx = 0

其他组合同样正交（同频率的正弦与余弦、不同频率的三角函数之间内积为零）。

系数公式

把 ff 写成

f(x)=a0+∑k=1∞(akcos⁡kx+bksin⁡kx)f(x)=a_0+\sum_{k=1}^{\infty}\left(a_k\cos kx+b_k\sin kx\right)

两边与 cos⁡x\cos x 做内积（即乘以 cos⁡x\cos x 再积分）。所有正交项消失，只剩

∫02πf(x)cos⁡x dx=a1∫02πcos⁡2x dx=a1π\int_0^{2\pi} f(x)\cos x\,dx = a_1\int_0^{2\pi}\cos^2x\,dx = a_1\pi

所以

a1=1π∫02πf(x)cos⁡x dxa_1=\frac{1}{\pi}\int_0^{2\pi} f(x)\cos x\,dx

一般地：

ak=1π∫02πf(x)cos⁡kx dx,bk=1π∫02πf(x)sin⁡kx dxa_k=\frac{1}{\pi}\int_0^{2\pi} f(x)\cos kx\,dx, \qquad b_k=\frac{1}{\pi}\int_0^{2\pi} f(x)\sin kx\,dx

常数项的系数是函数的平均值：

a0=12π∫02πf(x) dxa_0=\frac{1}{2\pi}\int_0^{2\pi} f(x)\,dx

一句话总结： 傅里叶系数就是函数在正交基上的投影坐标，公式与有限维的 xi=qiTvx_i=q_i^Tv 完全同构。

核心回顾

  • Markov 矩阵：非负、列和为 1；幂仍然是 Markov 矩阵。
  • 列和为 1 保证 A−IA-I 奇异，因此 1 必是特征值。
  • 其余特征值模小于 1，故 AkA^k 收敛到 λ=1\lambda=1 的特征向量方向。
  • 稳态向量非负，由初始总量确定倍数。
  • 正交基下，展开系数是一次内积：xi=qiTvx_i=q_i^Tv。
  • 函数空间的内积是积分；傅里叶系数就是投影坐标。

常见误区

  1. 把列和为 1 记成行和为 1。取决于向量写成列还是行；本课程用列。
  2. 认为所有 Markov 矩阵都收敛到唯一稳态。若存在模为 1 的其他特征值，可能周期振荡；若矩阵可约，稳态可能不唯一。
  3. 把 λ=1\lambda=1 的特征向量直接当成任意倍数。倍数由总人口（初始条件）决定。
  4. 把函数内积写成普通乘积。必须积分。
  5. 忘记傅里叶系数的归一化因子。cos⁡\cos、sin⁡\sin 的平方在 [0,2π][0,2\pi] 上积分是 π\pi，不是 1。

理解检验

1. 证明 Markov 矩阵的列和为 1 蕴含 A−IA-I 奇异。

参考答案：记 1=(1,1,…,1)T\mathbf{1}=(1,1,\dots,1)^T。列和为 1 等价于 1TA=1T\mathbf{1}^TA=\mathbf{1}^T（AA 的每一列与 1\mathbf{1} 的内积都是 1）。于是

1T(A−I)=1TA−1T=1T−1T=0.\mathbf{1}^T(A-I)=\mathbf{1}^TA-\mathbf{1}^T=\mathbf{1}^T-\mathbf{1}^T=0 .

1≠0\mathbf{1}\ne0 属于 A−IA-I 的左零空间，所以 A−IA-I 奇异，即 det⁡(A−I)=0\det(A-I)=0，λ=1\lambda=1 是特征值。也可以直接看：A−IA-I 的每列之和为 0，把它的各行相加得到零行，故矩阵奇异。

2. 为什么全 1 向量在 ATA^T 的零空间而不是 AA 的零空间？

参考答案：上面的关系是 1TA=1T\mathbf{1}^TA=\mathbf{1}^T，转置得 AT1=1A^T\mathbf{1}=\mathbf{1}。所以 1\mathbf{1} 是 ATA^T 的特征向量（特征值 1），也就是 (AT−I)1=0(A^T-I)\mathbf{1}=0，位于 ATA^T 的零空间。对 AA 本身，λ=1\lambda=1 的特征向量是稳态分布 x1x_1（分量非负、按总人口缩放），一般不是全 1 向量；只有行和也全为 1 的特殊矩阵（如双随机矩阵）才有 A1=1A\mathbf{1}=\mathbf{1}。行列和条件是两个不同的条件。

3. 在上面的两州模型中，若初始为 (1000,0)(1000,0)，稳态是什么？

参考答案：特征向量仍为 x1=(2,1)x_1=(2,1)、x2=(1,−1)x_2=(1,-1)。设

[10000]=c1[21]+c2[1−1],\begin{bmatrix}1000\\0\end{bmatrix}=c_1\begin{bmatrix}2\\1\end{bmatrix}+c_2\begin{bmatrix}1\\-1\end{bmatrix},

得 2c1+c2=10002c_1+c_2=1000、c1−c2=0c_1-c_2=0，所以 c1=c2=10003c_1=c_2=\frac{1000}{3}。当 k→∞k\to\infty，(0.7)k→0(0.7)^k\to0，稳态为

u∞=c1x1=[2000/31000/3].u_\infty=c_1x_1=\begin{bmatrix}2000/3\\1000/3\end{bmatrix}.

与初始在麻省的情形相同：稳态只由总人口 1000 决定，与初始分布无关。

4. 用内积推导 b1b_1 的公式。

参考答案：把 f(x)=a0+∑k≥1(akcos⁡kx+bksin⁡kx)f(x)=a_0+\sum_{k\ge1}(a_k\cos kx+b_k\sin kx) 两边乘以 sin⁡x\sin x，在 [0,2π][0,2\pi] 上积分。由正交性，除 b1sin⁡xb_1\sin x 外所有项积分为零（∫sin⁡xcos⁡kx=0\int\sin x\cos kx=0，∫sin⁡xsin⁡kx=0\int\sin x\sin kx=0 当 k≠1k\ne1）。剩下

∫02πf(x)sin⁡x dx=b1∫02πsin⁡2x dx=b1π,\int_0^{2\pi}f(x)\sin x\,dx=b_1\int_0^{2\pi}\sin^2x\,dx=b_1\pi,

所以 b1=1π∫02πf(x)sin⁡x dxb_1=\frac{1}{\pi}\int_0^{2\pi}f(x)\sin x\,dx。关键归一化：∫02πsin⁡2x dx=π\int_0^{2\pi}\sin^2x\,dx=\pi，不是 1。

5. 为什么说傅里叶级数是「无限维的正交基展开」？

参考答案：把函数看成向量空间中的元素，内积定义为 ⟨f,g⟩=∫02πfg dx\langle f,g\rangle=\int_0^{2\pi}fg\,dx。基函数 1,cos⁡x,sin⁡x,cos⁡2x,sin⁡2x,…1,\cos x,\sin x,\cos2x,\sin2x,\dots 两两正交，且每个的模长平方为 π\pi（常数 1 为 2π2\pi）。因此系数就是投影：

ak=⟨f,cos⁡kx⟩⟨cos⁡kx,cos⁡kx⟩,bk=⟨f,sin⁡kx⟩⟨sin⁡kx,sin⁡kx⟩,a_k=\frac{\langle f,\cos kx\rangle}{\langle\cos kx,\cos kx\rangle},\qquad b_k=\frac{\langle f,\sin kx\rangle}{\langle\sin kx,\sin kx\rangle},

与有限维的 xi=qiTvx_i=q_i^Tv 完全同构；区别只是基无穷多、求和变成级数，收敛性需要额外讨论（平方可积意义下成立）。

与前后课程的关系

这一讲把特征值用于概率模型，把投影用于函数空间。下一讲（复习课）串联正交、行列式与特征值三条主线；再之后进入对称矩阵与正定性，那里正交性、特征值与二次型会汇合。

What a Markov matrix is

A matrix is a Markov matrix if

  1. every entry is nonnegative, and
  2. every column sums to one.

In a row-vector convention the second condition becomes "every row sums to one". This course uses column vectors, so columns sum to one.

A typical example from the lecture:

A=[0.10.010.30.20.990.30.700.4]A=\begin{bmatrix} 0.1&0.01&0.3\\ 0.2&0.99&0.3\\ 0.7&0&0.4 \end{bmatrix}

Each column adds to 1. Both properties survive matrix multiplication, so every power of AA is again a Markov matrix.

Timeline

TimeTopic
0:00Markov matrices and Fourier series
3:05Recap: an example with λ=0\lambda=0
6:45Decomposing uku_k into λk\lambda^k terms
9:56Eigenvalue 1 and the decaying terms
12:57Markov matrices: columns sum to 1
16:01Dependent columns give a left nullvector
19:05Left eigenvectors and the steady state
22:29A population-mixing matrix
25:40The long-run distribution
28:46The second eigenvalue 0.7
31:48The full decomposition of uku_k
35:10Turning to projections and Fourier
38:13Expanding in an orthonormal basis
41:25Treating functions as vectors
44:30The Fourier coefficients
47:36A worked coefficient example

The key facts

FactStatement
1λ=1\lambda=1 is always an eigenvalue
2Every other eigenvalue satisfies $
3The eigenvector for λ=1\lambda=1 has no negative entries and gives the steady state

Why 1 must be an eigenvalue

Consider A−IA-I. Since each column of AA sums to 1, each column of A−IA-I sums to 0. Adding the rows of A−IA-I therefore produces the zero row, so the rows are dependent and the matrix is singular:

A−I singular  ⟺  det⁡(A−I)=0  ⟺  λ=1 is an eigenvalueA-I \text{ singular} \iff \det(A-I)=0 \iff \lambda=1 \text{ is an eigenvalue}

Another way to say it: the all-ones vector 1\mathbf{1} satisfies

1T(A−I)=0\mathbf{1}^T(A-I)=0

so 1\mathbf{1} lies in the left null space of A−IA-I. It is not an eigenvector of AA but of ATA^T. Since AA and ATA^T have the same eigenvalues — det⁡(AT−λI)=det⁡(A−λI)\det(A^T-\lambda I)=\det(A-\lambda I) — this also proves that 1 is an eigenvalue of AA.

Where the steady state comes from

The solution of uk+1=Auku_{k+1}=Au_k is

uk=c11kx1+c2λ2kx2+⋯u_k=c_1 1^k x_1+c_2\lambda_2^k x_2+\cdots

When every ∣λi∣<1|\lambda_i|<1, all terms except the first decay, so

uk→c1x1u_k\to c_1x_1

The steady state is the eigenvector for λ=1\lambda=1, scaled by a coefficient fixed by the initial total.

A population example

Two states, California (C) and Massachusetts (M). Each year:

  • 90% of California stays, 10% moves to Massachusetts;
  • 80% of Massachusetts stays, 20% moves to California.
[uCuM]k+1=[0.90.20.10.8][uCuM]k\begin{bmatrix}u_C\\u_M\end{bmatrix}_{k+1} = \begin{bmatrix}0.9&0.2\\0.1&0.8\end{bmatrix} \begin{bmatrix}u_C\\u_M\end{bmatrix}_{k}

Each column sums to 1, so the total population is conserved.

Eigenvalues

The trace is 1.71.7, so the other eigenvalue is 0.70.7:

λ1=1,λ2=0.7\lambda_1=1,\qquad \lambda_2=0.7

Eigenvectors

For λ=1\lambda=1, solve (A−I)x=0(A-I)x=0 with

A−I=[−0.10.20.1−0.2]A-I=\begin{bmatrix}-0.1&0.2\\0.1&-0.2\end{bmatrix}

whose null space is spanned by (2,1)(2,1).

For λ=0.7\lambda=0.7, solve (A−0.7I)x=0(A-0.7I)x=0 to get (1,−1)(1,-1).

A concrete trajectory

Start with u0=(0,1000)u_0=(0,1000): everyone in Massachusetts. Expand it in eigenvectors:

[01000]=c1[21]+c2[1−1]\begin{bmatrix}0\\1000\end{bmatrix} = c_1\begin{bmatrix}2\\1\end{bmatrix} +c_2\begin{bmatrix}1\\-1\end{bmatrix}

Solving gives

c1=10003,c2=−20003c_1=\frac{1000}{3},\qquad c_2=-\frac{2000}{3}

so

uk=10003[21]−20003(0.7)k[1−1]u_k=\frac{1000}{3}\begin{bmatrix}2\\1\end{bmatrix} -\frac{2000}{3}(0.7)^k\begin{bmatrix}1\\-1\end{bmatrix}

As k→∞k\to\infty the second term vanishes:

u∞=[2000/31000/3]u_\infty=\begin{bmatrix}2000/3\\1000/3\end{bmatrix}

The total is still 1000. A one-step check: u1=Au0=(200,800)u_1=Au_0=(200,800), matching the formula.

Background: This model assumes fixed yearly transition probabilities and no births, deaths, or migration outside the two states. A realistic population model would not conserve the total, and the matrix would not be Markov.

Projection onto an orthonormal basis

Let q1,…,qnq_1,\dots,q_n be an orthonormal basis of Rn\mathbb{R}^n. Any vector expands as

v=x1q1+x2q2+⋯+xnqnv=x_1q_1+x_2q_2+\cdots+x_nq_n

Take the inner product of both sides with qiq_i. Orthogonality kills every term except the ii-th:

qiTv=xi(qiTqi)=xiq_i^Tv=x_i(q_i^Tq_i)=x_i

Therefore

xi=qiTvx_i=q_i^Tv

In matrix form, Qx=vQx=v and Q−1=QTQ^{-1}=Q^T, so x=QTvx=Q^Tv. Orthonormality turns finding coefficients into a single inner product.

From vectors to functions: Fourier series

Fourier's insight is to carry this structure into function space:

  • the vector vv becomes a function f(x)f(x);
  • the basis vectors become the functions 1,cos⁡x,sin⁡x,cos⁡2x,sin⁡2x,…1,\cos x,\sin x,\cos 2x,\sin 2x,\dots;
  • finite dimension becomes infinite dimension.

The inner product of functions

A vector inner product multiplies components and adds. A function has a continuum of points, so the sum becomes an integral:

⟨f,g⟩=∫02πf(x)g(x) dx\langle f,g\rangle=\int_0^{2\pi} f(x)g(x)\,dx

The interval is [0,2π][0,2\pi] because every basis function has period 2π2\pi.

Orthogonality

For sin⁡x\sin x and cos⁡x\cos x:

∫02πsin⁡xcos⁡x dx=12∫02πsin⁡2x dx=0\int_0^{2\pi}\sin x\cos x\,dx = \frac12\int_0^{2\pi}\sin 2x\,dx = 0

Other pairs are orthogonal as well: same-frequency sine and cosine, and different frequencies, all integrate to zero.

The coefficient formulas

Write

f(x)=a0+∑k=1∞(akcos⁡kx+bksin⁡kx)f(x)=a_0+\sum_{k=1}^{\infty}\left(a_k\cos kx+b_k\sin kx\right)

Take the inner product with cos⁡x\cos x, meaning multiply by cos⁡x\cos x and integrate. Every orthogonal term vanishes and only one survives:

∫02πf(x)cos⁡x dx=a1∫02πcos⁡2x dx=a1π\int_0^{2\pi} f(x)\cos x\,dx = a_1\int_0^{2\pi}\cos^2x\,dx = a_1\pi

Hence

a1=1π∫02πf(x)cos⁡x dxa_1=\frac{1}{\pi}\int_0^{2\pi} f(x)\cos x\,dx

In general,

ak=1π∫02πf(x)cos⁡kx dx,bk=1π∫02πf(x)sin⁡kx dxa_k=\frac{1}{\pi}\int_0^{2\pi} f(x)\cos kx\,dx, \qquad b_k=\frac{1}{\pi}\int_0^{2\pi} f(x)\sin kx\,dx

The constant term is the average value:

a0=12π∫02πf(x) dxa_0=\frac{1}{2\pi}\int_0^{2\pi} f(x)\,dx

One-sentence summary: Fourier coefficients are the coordinates of a function in an orthonormal basis, the exact analogue of xi=qiTvx_i=q_i^Tv.

Core review

  • A Markov matrix is nonnegative with columns summing to one; its powers are Markov too.
  • Column sums of one make A−IA-I singular, so 1 is an eigenvalue.
  • Every other eigenvalue has modulus below 1, so AkA^k converges to the λ=1\lambda=1 direction.
  • The steady-state vector is nonnegative, scaled by the initial total.
  • In an orthonormal basis, coefficients are single inner products: xi=qiTvx_i=q_i^Tv.
  • Function inner products are integrals; Fourier coefficients are projection coordinates.

Common misconceptions

  1. Remembering "rows sum to one" instead of columns. It depends on the vector convention; this course uses columns.
  2. Assuming every Markov matrix converges to a unique steady state. Other eigenvalues of modulus 1 can produce oscillation, and a reducible matrix may have several steady states.
  3. Treating the λ=1\lambda=1 eigenvector as arbitrary up to scale. The scale is fixed by the initial total.
  4. Writing a function inner product as a plain product. It must be integrated.
  5. Forgetting the normalization in Fourier formulas. The integral of cos⁡2\cos^2 over [0,2π][0,2\pi] is π\pi, not 1.

Check your understanding

1. Prove that column sums of one force A−IA-I to be singular.

Reference answer: Write 1=(1,1,…,1)T\mathbf{1}=(1,1,\dots,1)^T. Column sums equal to 1 mean 1TA=1T\mathbf{1}^TA=\mathbf{1}^T (each column has inner product 1 with 1\mathbf{1}). Hence

1T(A−I)=1TA−1T=1T−1T=0.\mathbf{1}^T(A-I)=\mathbf{1}^TA-\mathbf{1}^T=\mathbf{1}^T-\mathbf{1}^T=0 .

Since 1≠0\mathbf{1}\ne0 lies in the left nullspace of A−IA-I, the matrix is singular: det⁡(A−I)=0\det(A-I)=0 and λ=1\lambda=1 is an eigenvalue. Equivalently, every column of A−IA-I sums to zero, so adding its rows gives a zero row.

2. Why is the all-ones vector in the null space of ATA^T rather than of AA?

Reference answer: The relation above is 1TA=1T\mathbf{1}^TA=\mathbf{1}^T; transposing gives AT1=1A^T\mathbf{1}=\mathbf{1}. So 1\mathbf{1} is an eigenvector of ATA^T with eigenvalue 1, i.e. (AT−I)1=0(A^T-I)\mathbf{1}=0, which places it in the nullspace of ATA^T. For AA itself, the eigenvector for λ=1\lambda=1 is the steady-state vector x1x_1 (nonnegative, scaled by total population), generally not the all-ones vector. Only matrices whose row sums are also 1 (doubly stochastic) satisfy A1=1A\mathbf{1}=\mathbf{1}; column and row sum conditions are different.

3. In the two-state model, what is the steady state if the initial population is (1000,0)(1000,0)?

Reference answer: The eigenvectors are still x1=(2,1)x_1=(2,1) and x2=(1,−1)x_2=(1,-1). Write

[10000]=c1[21]+c2[1−1],\begin{bmatrix}1000\\0\end{bmatrix}=c_1\begin{bmatrix}2\\1\end{bmatrix}+c_2\begin{bmatrix}1\\-1\end{bmatrix},

so 2c1+c2=10002c_1+c_2=1000 and c1−c2=0c_1-c_2=0, giving c1=c2=10003c_1=c_2=\frac{1000}{3}. As k→∞k\to\infty, (0.7)k→0(0.7)^k\to0, and the steady state is

u∞=c1x1=[2000/31000/3].u_\infty=c_1x_1=\begin{bmatrix}2000/3\\1000/3\end{bmatrix}.

It matches the case that started entirely in Massachusetts: the steady state depends only on the total population 1000, not on the initial distribution.

4. Derive the formula for b1b_1 using an inner product.

Reference answer: Multiply f(x)=a0+∑k≥1(akcos⁡kx+bksin⁡kx)f(x)=a_0+\sum_{k\ge1}(a_k\cos kx+b_k\sin kx) by sin⁡x\sin x and integrate over [0,2π][0,2\pi]. Orthogonality kills everything except b1sin⁡xb_1\sin x (∫sin⁡xcos⁡kx=0\int\sin x\cos kx=0 and ∫sin⁡xsin⁡kx=0\int\sin x\sin kx=0 for k≠1k\ne1). What remains is

∫02πf(x)sin⁡x dx=b1∫02πsin⁡2x dx=b1π,\int_0^{2\pi}f(x)\sin x\,dx=b_1\int_0^{2\pi}\sin^2x\,dx=b_1\pi,

hence b1=1π∫02πf(x)sin⁡x dxb_1=\frac{1}{\pi}\int_0^{2\pi}f(x)\sin x\,dx. The normalisation ∫02πsin⁡2x dx=π\int_0^{2\pi}\sin^2x\,dx=\pi (not 1) is essential.

5. In what sense is a Fourier series an expansion in an infinite orthonormal basis?

Reference answer: Treat functions as vectors with inner product ⟨f,g⟩=∫02πfg dx\langle f,g\rangle=\int_0^{2\pi}fg\,dx. The functions 1,cos⁡x,sin⁡x,cos⁡2x,sin⁡2x,…1,\cos x,\sin x,\cos2x,\sin2x,\dots are pairwise orthogonal, with squared norms π\pi (and 2π2\pi for the constant 1). The coefficients are therefore projections,

ak=⟨f,cos⁡kx⟩⟨cos⁡kx,cos⁡kx⟩,bk=⟨f,sin⁡kx⟩⟨sin⁡kx,sin⁡kx⟩,a_k=\frac{\langle f,\cos kx\rangle}{\langle\cos kx,\cos kx\rangle},\qquad b_k=\frac{\langle f,\sin kx\rangle}{\langle\sin kx,\sin kx\rangle},

exactly analogous to xi=qiTvx_i=q_i^Tv in finite dimensions. The only differences are that the basis is infinite and the sum becomes a series, whose convergence needs a separate argument (it holds in the square-integrable sense).

How this connects

This lecture applies eigenvalues to probability models and projection to function spaces. The next session reviews orthogonality, determinants, and eigenvalues together; after that the course turns to symmetric matrices and positive definiteness, where orthogonality, eigenvalues, and quadratic forms meet.

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